Chapter 1
Introduction
Chapter Abstract. In this introductory chapter we first discuss the magnificent achievements
in aviation during the past century, which is the most obvious area in which fluid mechanics
plays a significant role. It is not the only area where fluid mechanics plays an important role
and we discuss several other applications.
Then, we focus on dimensions and units commonly used in fluid mechanics. Units and
dimensions allow us to quantify physical properties such as position, velocity, volume and
mass. Next, we define a fluid and introduce the concept of the continuum approximation,
which permits us to relate what occurs in a fluid at the microscopic level to the macroscopic
behavior that we observe in everyday applications. We have a quick look at some of the
thermodynamic properties of fluids, including specialized properties such as vapor pressure,
surface tension, compressibility and viscosity. The chapter concludes with the definitions of
a real fluid and an ideal fluid along with an overview of how the science of fluid mechanics
has evolved.
By and large, all branches of engineering use the same mathematical tools and laws of
physics, so that the primary distinguishing feature is nomenclature. Thus, one of the most
important objectives of this chapter is to introduce fluid-mechanics nomenclature.
1.1 Aiming for the Stars
There is little doubt that from the days when humans gazed up at the sky above the warm
savannas and cold caves they marveled at the ability of birds to fly. And in the evenings
they surely wondered about the lights that dotted the sky. It cannot be otherwise for, as the
philosopher Aristotle said, “All men by nature desire to know.”
The invention of the telescope provided the first great advance toward discovering the
nature of the universe. But, it provided no clues about how mankind might travel beyond the
confines of the Earth and its atmosphere.
The twentieth century ushered in the era of flight. The century had barely begun when, on
December 17, 1903, Orville and Wilbur Wright piloted the first heavier-than-air craft to fly
under its own power. Named Wright Flyer I, it was the result of several achievements by the
Wright brothers, including: extensive and painstaking wind-tunnel tests; a new lightweight
engine design; and an efficient propeller. While their first flight lasted only about 12 seconds,
it proved that powered flight was possible.
1
2
CHAPTER 1. INTRODUCTION
Figure 1.1: With their historic flight at Kitty Hawk, NC on December 17, 1903, Orville and
Wilbur Wright accomplished a long-sought goal of mankind—flight. [Photographs courtesy
of and c www.arttoday.com]
Orville was the pilot of the first flight, and he lay next to the motor on the lower wing.
Wilbur steadied the vehicle at one of the wing tips. After running along the ground for about
40 feet, the Wright Flyer, shown in Figure 1.1, lifted off of a coastal sand dune near Kitty
Hawk, North Carolina. In the 12 seconds before it returned to the ground, the airplane had
flown 120 feet. They made several flights on that historic day, with Wilbur piloting the
longest—852 feet in 59 seconds.
Since that historic day, aviation has brought us faster and more efficient vehicles that
permit mankind to “soar with the eagles.” Civil aviation developed rapidly during the next 40
years and by the end of World War II in 1945, estimates indicate that at least a million people
had flown in an airplane.
Just 44 years after the Wright brothers’ first powered flight, United States Air Force test
pilot Chuck Yeager demonstrated that it is possible to fly at speeds in excess of the speed of
sound, an issue that was very much in doubt at the time. He was selected from a group of
volunteers to test the experimental X-1 aircraft, which is depicted in Figure 1.2. The goal of
the X-1 flight was to determine whether or not an airplane could fly faster than the speed of
sound, and if a pilot could control the plane despite the effects of shock waves.
Figure 1.2: Chuck Yeager broke through the sound barrier with his October 14, 1947 flight of
the X-1—nicknamed the “Glamorous Glennis”—over Rogers Dry Lake in California. [Photographs courtesy of the U. S. Air Force]
1.1. AIMING FOR THE STARS
3
The flight took place on October 14, 1947 above Rogers Dry Lake in California. The X-1
was launched from a B-29 mother ship at an altitude of 25000 feet. Yeager then flew the
rocket-powered X-1 to an altitude of 43000 feet. He succeeded in flying at a Mach number
of 1.06 with the X-1 remaining intact, and thus became the first person to break the sound
barrier. Yeager’s rocket-powered flight is often cited as the beginning of the space age.
The development of rockets has taken us beyond the confines of Earth’s gravitational field.
Those of us old enough to be tuned in, on color television broadcasting live from the Moon
(an achievement in its own right), still remember the chills up and down our spine on July
20, 1969, when Neil Armstrong (Figure 1.3) said, as he became the first man to set foot on
the Moon, “That’s one small step for man, one giant leap for mankind.”
Figure 1.3: On July 20, 1969, Neil Armstrong became the first human to set foot on the moon.
[Photographs courtesy of NASA]
Today, we see routine flights of manned space vehicles performing missions in Earth
orbit and rocket launches from all over the globe. In particular, there have been numerous
private-sector commercial achievements in space, most notably in the area of communications
satellites.
Nevertheless, manned space travel has remained an unrealized revolution. In contrast to
what occurred in the 40 years after the Wright brothers’ flight, in a comparable time period
since the first humans rocketed into space, fewer than 500 individuals have followed them.
At the dawn of the twenty-first century, another historic milestone has been reached that
holds promise to advance mankind’s desire for space travel and exploration. On October 4,
2004, the United States of America’s private-enterprise system demonstrated its vast powers of
creativity and can-do attitude in a most dramatic way. Scaled Composites, an entrepreneurial
venture founded by aviation pioneer Burt Rutan, launched a vehicle named SpaceShipOne for
the second time in two weeks. SpaceShipOne rocketed to an altitude of 69.6 miles above the
Earth’s surface and captured the ten-million dollar Ansari X-Prize.
As with Chuck Yeager’s 1947 flight, SpaceShipOne was launched from a mother ship.
The mother ship, named the White Knight, took off from the Mojave Airport and carried
SpaceShipOne to an altitude of 47000 feet. After detaching from the mother ship, SpaceShipOne then fired its rocket engine for 84 seconds, which propelled the vehicle to a speed in
excess of Mach 3. At the maximum altitude, pilot Brian Binnie was weightless for three and
a half minutes. To re-enter the Earth’s atmosphere, SpaceShipOne’s wings were folded into
a self-stabilizing, high-drag configuration. After re-entry, its wings were folded back into a
glider configuration for landing.
4
CHAPTER 1. INTRODUCTION
Figure 1.4: On October 4, 2004, entrepreneur Burt Rutan’s SpaceShipOne rocketed into
suborbital space, 69.6 miles above the Earth. [Photographs courtesy of Scaled Composites]
The saga of SpaceShipOne, shown in Figure 1.4, is punctuated with salutes to aviation
history. For example, the private-sector Ansari X-Prize is modeled after the Orteg Prize that
Charles Lindbergh won for his 1927 solo flight across the Atlantic Ocean. Brian Binnie also
had the distinction of piloting SpaceShipOne’s first powered, supersonic flight on November
17, 2003, the one-hundredth anniversary of the Wright brothers’ flight.
Scaled Composites has teamed with Virgin Group to create The Spaceship Company. This
new aerospace company has built SpaceShipTwo, depicted in Figure 1.5, which is the first
vehicle of “the world’s first fleet of commercial spaceships and launch aircraft.” There were
sixteen test flights between October 2010 and September 2011, and SpaceShipTwo performed
in a near-perfect manner throughout testing. The vehicle will carry six passengers and two
pilots to an altitude 68 miles above Earth. Like SpaceShipOne it launches from a mother ship
named White Knight Two. Virgin Atlantic plans on operating a fleet of five of these spacecraft
to carry passengers into space beginning in 2013. Needless to say, the entrance of private
enterprise into manned space travel puts an awesome new participant in quest of mankind’s
long-sought dream of traveling to the heavens.
Figure 1.5: An artist’s conception of Virgin Galactic’s SpaceShipTwo. [Photograph courtesy
of Virgin Galactic]
1.1. AIMING FOR THE STARS
5
All of these achievements have been accomplished partly because of the vast accumulation
of knowledge regarding the motion of fluids. The lifting and propulsive forces generated by
the Wright Flyer have been thoroughly understood, and dramatically improved designs have
evolved for modern aircraft. Theoretical studies of flows at supersonic speeds were conducted
by German researcher Ludwig Prandtl in the early part of the twentieth century, long before
supersonic flight was even dreamed of. Those studies provided part of the basis for Chuck
Yeager’s historic flight. Intense theoretical and experimental analysis in the 1950’s and 1960’s
provided accurate predictions of the heating that a space vehicle would have to be protected
against to permit safe reentry into the Earth’s atmosphere. The emergence of Computational
Fluid Dynamics (CFD) has put the computer at our disposal to aid in the design of advanced
vehicles such as the Space Shuttle and SpaceShipOne.
While evidence of its importance and relevance to real-world problems is most obvious
in the achievements with aircraft and rockets, fluid mechanics is not limited solely to such
problems. The aerodynamic drag of automobiles early in the twenty-first century is less
than 40% of the drag of models designed in the early 1900’s (cf. Figure 1.6). The flow of
gases through the engine of an automobile and even its air-conditioning system are routinely
computed by today’s auto builders.
Figure 1.6: Left: 1914 sedan with a drag coefficient of 0.80. Right: 2002 Corvette with a
drag coefficient of 0.29. [Corvette photograph courtesy of William Watts.]
A challenging area of fluid-mechanics research is the flow of blood through arteries and,
ultimately, through the human heart. Developing a thorough understanding of blood flow
through the body will surely help control and/or eliminate some of the most serious health
hazards in today’s world such as arteriosclerosis and heart disease.
Applications involving fluid motion extend to all branches of engineering. Aeronautical
and mechanical engineers’ interests range from basic studies of fluid motion, most notably
turbulence, to everyday problems involving power generation, heating and ventilation, computer disk-drive design, etc. Civil engineers focus on interaction of aerodynamic forces with
structures such as bridges and piers. Electrical engineers seek reduced costs in forming microchips, where acids must flow in a controlled manner to create desired patterns on silicon and
other semiconductor materials. Chemical engineers must accurately determine reaction rates,
a particularly acute problem when the velocity is high enough for the flow to be turbulent.
This book has been written with the object of introducing the reader to the exciting field
of fluid mechanics. The book strives for mathematical rigor throughout, without concealing
the fundamental physical concepts involved. So, read on with the understanding that the intent
is to challenge your intellect, and to encourage you to hone your skills in mathematical and
physical reasoning. Aiming high is a good strategy to follow if you truly want to reach for
the stars!
6
CHAPTER 1. INTRODUCTION
1.2 Dimensions and Units
Like all of the branches of physical science, the field of fluid mechanics has developed through
a balance between theory and experiment. In order to have confidence in any theoretical
development, we must demand that the quantitative predictions of the theory are consistent
with, and within the accuracy of, measurements.
To describe physical properties in quantitative terms, we must select a system of units for
expressing their dimensions. There are several systems of units used in general engineering
practice, and this section describes four of the commonly used systems. Of these four, this
text almost exclusively uses Standard International units and one of the subsets of the U. S.
Customary System of units. In the text, we refer to these two sets of units as SI and USCS,
respectively. Because of their inherent simplicity, SI units are becoming the standard in many
branches of engineering. Several engineering journals, for example, require authors who use
USCS units to parenthetically include corresponding values in SI units.
1.2.1 Independent Dimensions
Dimensions denote physical properties in a generic sense independent of the units we select.
The first thing we must decide upon is what dimensions are independent of all others, i.e.,
which are most basic in some sense. Just as the spatial coordinates x, y and z are independent
variables in a three-dimensional rectangular Cartesian coordinate system, so we must choose
a set of independent dimensions. Any other dimensions are then expressed in terms of the
independent dimensions, and are called secondary dimensions or dependent dimensions.
For engineering work, we select either mass, length, time and temperature or force, length,
time and temperature as the independent dimensions. In either case, Newton’s second law
of motion provides the defining relationship between mass and force, regardless of which is
chosen as the independent dimension. We will return to the concepts of independent and
secondary dimensions in Chapter 2.
1.2.2 Common Systems of Units
Various groups and countries have established systems of units and associated independent
dimensions. Part of the objective in establishing these standards has been to promote the use
of consistent units. This means no special conversion factors are required in a physicallybased equation to make it dimensionally homogeneous, i.e., to make all terms in the equation
have the same dimensions. Three of the four systems discussed in this section satisfy this
constraint. As a counter example, one of the systems, viz., the English Engineering system,
violates this objective with regard to Newton’s second law of motion.
The two most prevalent types of units are Metric Units and the U. S. Customary System
(USCS). Each has two sub-types. In the case of Metric Units, there is the Standard International (SI) system and the Centimeter-Gram-Second (CGS) system. Table 1.1 lists mass,
length, time, temperature and force in the SI and CGS systems of units.
For the USCS, the sub-types are known as the British Gravitational (BG) system and the
English Engineering (EE) system. Table 1.2 lists mass, length, time, temperature and force
in the BG and EE systems of units. This text exclusively uses either the SI system or the
BG subset of the U. S. Customary System. Because of its unusual nature, the text completely
avoids the EE system.
1.2. DIMENSIONS AND UNITS
7
Table 1.1: Metric Units
Dimension
Mass
Length
Time
Temperature
Force
Standard International (SI)
kilogram (kg)
meter (m)
second (sec)
Kelvin (K)
Newton (N)
Centimeter-Gram-Second (CGS)
gram (g)
centimeter (cm)
second (sec)
Kelvin (K)
dyne (dyne)
Table 1.2: U. S. Customary System of Units (USCS)
Dimension
Mass
Length
Time
Temperature
Force
British Gravitational (BG)
slug (slug)
foot (ft)
second (sec)
o
Rankine (o R)
pound (lb)
English Engineering (EE)
pound-mass (lbm)
foot (ft)
second (sec)
o
Rankine (o R)
pound-force (lbf)
Standard International Units
The independent dimensions in the SI system1 are as follows. The unit of mass is the kilogram
(kg), the unit of length is the meter (m), the unit of time is the second (sec) and the unit of
temperature is the Kelvin (K). The temperature expressed in Kelvins is absolute temperature,
and is related to the commonly used Celsius, or centigrade, scale (o C) as follows.
K=
o
C + 273.15
(1.1)
The most important secondary dimension for applications using the laws of physics is force.
The unit of force is called the Newton (N), which is defined as the force required to give a
mass of one kilogram an acceleration of one meter per second per second. In dimensional
terms, we say
1 N = (1 kg) · (1 m/sec2 ) = 1 kg · m/sec2
(1.2)
Note that, since the acceleration of gravity is 9.807 m/sec2 , this means the weight of a onekilogram mass is 9.807 Newtons. Note also that when—in common parlance—the “weight”
of an object is expressed in kilograms, what is actually being quantified is the object’s mass.
Two other important secondary dimensions are those of work and power. In the SI system,
the units of work and power are the Joule (J) and the Watt (W), respectively. A Joule is the
amount of work done when a one-Newton force produces a displacement of one meter in the
direction of the force. A Watt is one Joule per second. Algebraically, we say
1 J=1 N·m
1 W = 1 J/sec = 1 N · m/sec
(1.3)
1 In older texts, this system is often referred to as the MKS system, which is an acronym for meter-kilogram-second.
Its modern name is also Système Internationale.
8
CHAPTER 1. INTRODUCTION
Centimeter-Gram-Second Units
In the CGS system, we select the units of mass, length, time and temperature to be the gram (g),
the centimeter (cm), the second (sec) and the Kelvin (K), respectively. The unit of force is
called the dyne (dyne), which is defined as the force required to give a mass of one gram an
acceleration of one centimeter per second per second. Hence,
1 dyne = (1 g) · (1 cm/sec2 ) = 1 g · cm/sec2 = 10−5 N
(1.4)
Because the acceleration of gravity is 980.7 cm/sec2 , this means the weight of a one-gram
mass is 980.7 dynes.
In the CGS system, the unit of work is the erg (erg), while power is expressed in terms
of ergs per second (erg/sec). An erg is defined as the amount of work done when a one-dyne
force produces a displacement of one centimeter in the direction of the force. Table 1.3 lists
the prefixes used with metric units.
Table 1.3: Metric-Unit Prefixes
Factor
1,000,000,000,000,000,000,000,000
1,000,000,000,000,000,000,000
1,000,000,000,000,000,000
1,000,000,000,000,000
1,000,000,000,000
1,000,000,000
1,000,000
1,000
100
10
0.1
0.01
0.001
0.000 001
0.000 000 001
0.000 000 000 001
0.000 000 000 000 001
0.000 000 000 000 000 001
0.000 000 000 000 000 000 001
0.000 000 000 000 000 000 000 001
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
1024
1021
1018
1015
1012
109
106
103
102
101
10−1
10−2
10−3
10−6
10−9
10−12
10−15
10−18
10−21
10−24
Prefix
Symbol
yotta
zetta
exa
peta
tera
giga
mega
kilo
hecto
deca
deci
centi
milli
micro
nano
pico
femto
atto
zepto
yocto
Y
Z
E
P
T
G
M
k
h
da
d
c
m
μ
n
p
f
a
z
y
British Gravitational Units
In the BG system, the independent dimensions are force, length, time and temperature. The unit
of force is the pound (lb), the unit of length is the foot (ft), the unit of time is the second (sec)
and the unit of temperature is the degree Rankine (o R). The temperature expressed in degrees
Rankine is absolute temperature, and is related to the Fahrenheit scale (o F) by
o
R=
o
F + 459.67o
(1.5)
1.2. DIMENSIONS AND UNITS
9
The unit of mass is called the slug (slug), which is defined as the mass upon which application
of a one-pound force will cause an acceleration of one foot per second per second. Thus,
2
2
(1.6)
1 lb = (1 slug) · (1 ft/sec ) = 1 slug · ft/sec
The acceleration of gravity in BG units is 32.174 ft/sec , so that a mass of one slug weighs
32.174 pounds. Finally, the units of work and power in the BG system are the British thermal
unit (Btu) and the horsepower (hp), respectively. A Btu is the amount of heat required to raise
the temperature of one pound of water (at its maximum density) by one degree Rankine. It is
equal to 1055 Joules. A horsepower is numerically equal to the rate of 33000 foot-pounds of
work per minute.2 It is equal to 0.7457 kiloWatts.
2
1 Btu = 778 ft · lb
(1.7)
1 hp = 2545 Btu/hr = 550 ft · lb/sec
English Engineering Units
In the EE system, the units for mass and force are defined separately. The unit of force is the
pound-force (lbf), while the unit of mass is the pound-mass (lbm). As with the BG system,
the units of length, time and temperature are the foot, the second and the degree Rankine,
respectively. Having separate units for mass and force adds the complication that Newton’s
second law of motion must be written as
ma
F=
(1.8)
gc
where F is force, m is mass, a is acceleration and gc is a dimensional constant determined
as follows. We define the pound-force and the pound-mass such that application of a one-lbf
force to a one-lbm mass yields an acceleration equal to the acceleration of gravity on Earth,
32.174 ft/sec2 . Hence,
2
1 lbf =
1 lbm · 32.174 ft/sec
gc
=⇒
gc = 32.174
ft · lbm
lbf · sec2
(1.9)
Because of the dimensional parameter gc appearing in a basic law of physics, EE units are
inconvenient, and find use mainly when convention dictates. The primary fields in which EE
units are found are thermodynamics and heat transfer.
o
Example 1.1 A nanotube researcher chooses to express length in Angstrom units (A), mass in
grams (g) and time in microseconds (μsec). In this system of units, infer the logical choice of units
for force.
o
Solution. First, we note that 1 A = 10−10 m, 1 g = 10−3 kg and 1 μsec = 10−6 sec. Since force
has units of mass times length divided by time squared, the units are
10−3 kg · 10−10 m
kg · m
= 10−1
= 10−1 N = 1 dN
(10−6 sec)2
sec2
where we observe from Equation (1.2) that one Newton is 1 kg · m/sec2 and, from Table 1.3, “dN”
denotes deciNewton. Thus, the logical choice of units for expressing force in this system would be
the deciNewton.
2 The term horsepower was originated by Boulton and Watt to state the power of their steam engines. In a practical
test, it was found that the average horse could work constantly at the rate of 22000 foot-pounds per minute. This
was increased by one half in defining this arbitrary, and now universal, unit of power.
10
CHAPTER 1. INTRODUCTION
1.2.3 Converting Dimensional Quantities
For the discussion to follow, it is convenient to identify the independent dimensions in algebraic
terms. We will refer to mass as M , length as L, time as T , temperature as Θ, and force as F .
As noted above, secondary dimensions are related to the independent dimensions by a physical
principle or by definition. For instance, the dimensional representation of velocity, U , is
[U ] =
L
T
(1.10)
where [U ] means dimensions of U . Similarly, pressure has dimensions F/L2 and acceleration is expressed dimensionally as L/T 2 . We will focus further on the distinction between
dimensions and units in Chapter 2.
Converting from one system of units to another is straightforward if we follow some
simple algebraic rules. For example, Table 1.4, which lists conversion factors for properties
commonly encountered in fluid mechanics, tells us that 1 pound is equal to 4.448 Newtons.
Also, 1 foot is 0.3048 meters. Hence, to convert from pressure expressed in lb/ft2 to N/m2 ,
we proceed as follows. We form the ratios of independent or secondary dimensions, and use
them as dimensional multiplicative factors with a magnitude of unity. That is, we say
4.448 N
1 lb
=1
and
0.3048 m
1 ft
=1
(1.11)
We can now use these factors in a strict algebraic sense to convert from the USCS system to
the SI system as follows.
1
lb
(1 lb) · (4.448 N/lb)
N
2 =
2 = 47.88 m2
2
ft
(1 ft ) · (0.3048 m/ft)
(1.12)
Note the precise algebraic cancellation of “lb” and “ft2 ” in the two factors after the first equals
sign. Organizing unit computations in this manner greatly reduces the chance of arithmetic
error in converting from one set of units to another.
Example 1.2 The units of energy in the SI system are Newton·meter. What is the conversion factor
from the SI system to foot·pounds in the USCS system?
Solution. Using Equation (1.11), we can say
1 ft · lb = (1 ft) · 0.3048
m
ft
· (1 lb) · 4.448
N
lb
= 1.3558 N · m
Both SI and USCS units are used in modern engineering applications, especially in the
aerospace industry. While there is a trend toward SI units, many important engineering journal
articles and books that are in common usage provide information in terms of the USCS. Such
articles and books continue to constitute important reference sources that reflect many decades
of engineering research and experimentation. Hence, the conscientious student will make a
point of being “bilingual” regarding these systems of units. In order to promote this end,
approximately half of the homework problems in this text are cast in terms of each of the two
predominant systems. To avoid unnecessary error, be sure to work the problems in the units
given. Converting to another system of units simply introduces potential arithmetic errors that
have nothing to do with mastering the physical principles involved in the problem.
1.2. DIMENSIONS AND UNITS
11
The homework problems throughout this book strive toward maximizing physical understanding while minimizing arithmetic error. Specifically, most of the problems require you to
first derive an algebraic result. Formulas and equations, after all, are the things that a practicing engineer incorporates into a computer program in order to conduct a design study. Also,
as any conscientious practicing engineer will tell you, the equations you derive will prove to
be especially helpful in assuring that your answer is algebraically correct to the extent that it
has the correct dimensions (Chapter 2 discusses this concept in great detail). To help familiarize you with typical engineering magnitudes for key flow properties, many of the problems
provide numerical values that can be substituted into the algebraic formula.
Table 1.4: Units Conversion Factors
Length
(L)
1
1
1
1
1
ft
in
mi
km
m
=
=
=
=
=
0.3048 m
25.4 mm
5280 ft
0.6214 mi
3.2808 ft
Velocity
(LT −1 )
1
1
1
1
1
mph
mph
knot
knot
m/sec
=
=
=
=
=
1.467
0.447
1.688
0.514
3.281
Area
(L2 )
1
1
1
1
ft2
in2
cm2
m2
=
=
=
=
0.0929
6.4516
0.1550
10.764
Rotation
(T −1 )
1
1
1
1
rpm
Hz
cps
rev
=
=
=
=
0.1047 sec−1
2π sec−1
2π sec−1
2π radians
Volume
(L3 )
1
1
1
1
gal
gal
gal
L
=
=
=
=
0.0037854 m3
231 in3
0.134 ft3
0.001 m3
Temperature
(Θ)
o
C
K
o
R
o
R
=
=
=
=
o
5 o
9 ( F−32 )
o
Pressure
(M L−1 T −2 )
1
1
1
1
1
psi
psf
atm
atm
atm
=
=
=
=
=
6.895 kPa
47.88 Pa
101 kPa
2116.8 psf
760 mmHg
Work
(M L2 T −2 )
1
1
1
1
1
ft·lb
Btu
Btu
J
J
=
=
=
=
=
1.3558 J
778 ft·lb
1055 J
0.239 cal
0.7376 ft·lb
Force
(M LT −2 )
1
1
1
1
1
oz
lb
ton
N
kN
=
=
=
=
=
0.278
4.448
8.897
3.597
224.8
Power
(M L2 T −3 )
1
1
1
1
hp
hp
kW
J/sec
=
=
=
=
550 ft·lb/sec
2545 Btu/hr
1.341 hp
1W
Mass
(M )
1 slug
1 kg
=
=
14.594 kg
0.0685 slug
Gravity
(LT −2 )
g
g
=
=
32.174 ft/sec2
9.807 m/sec2
m2
cm2
in2
ft2
N
N
kN
oz
lb
ft/sec
m/sec
ft/sec
m/sec
ft/sec
C + 273.15
9
K
5
o
F + 459.67o
12
CHAPTER 1. INTRODUCTION
Example 1.3 If you are driving along a highway at a speed of 53 mph, what is your speed in
Astronomical Units per Millennium (AUpM)?
Solution. By definition, the Astronomical Unit (AU) is the average distance between the Earth and
the sun, which is 93 · 106 miles. Also, a millennium (Mln) is 1000 years, each of which has 365.25
days that are 24 hours long. Hence, to convert from mph to AUpM, we first note that
1 Mln = (1000 yr) 365.25
day
yr
24
hr
day
= 8.766 · 106 hr
Therefore,
93 · 106 mi/AU (1 AU)
= 10.6 mph
(8.766 · 106 hr/Mln) (1 Mln)
So, your speed of 53 mph expressed in Astronomical Units per millennium is
1 AUpM =
Speed =
(53 mph)
= 5 AUpM
(10.6 mph/AUpM)
1.3 Definition of a Fluid
To begin our study, we first note that the field we call fluid mechanics is the branch of
physics concerned with fluid motion. Most substances found in nature can be thought of as
either a solid or a fluid, where the two most common fluids are gases and liquids. The goals
of this book are twofold. Our most important goal is to explain how to apply Newton’s laws
of motion to fluids. Our second goal is to present a brief overview of the general field of fluid
mechanics. The first thing we need to do is define what we mean by a fluid.
1.3.1 Simplistic Definition of a Fluid
Anything that flows. Liquids and gases are the most obvious examples of substances that
will flow. Traffic (Figure 1.7) is a more subtle example that can be treated as a fluid,
in the sense that its motion can be described with equations derived from fluid-mechanics
principles. The flow of traffic has been modeled [Lighthill and Whitham (1955)] as a fluid
whose density (number of automobiles per unit distance) varies with speed. A particularly
interesting application of traffic-flow theory is to the timing of traffic lights. In the 1960’s
and 1970’s, many cities posted signs indicating the speed required to avoid catching a red
light on a given route. The speed and timing of the lights were determined from observations
of normal traffic flow and the Lighthill-Whitham theory.
1.3.2 Rigorous Definition of a Fluid
A substance that cannot be in static equilibrium under the action of an oblique stress.3
On the one hand, if only normal forces such as pressure act, a fluid will adjust to the applied
pressure with a change in volume. After adjusting, it remains at rest as illustrated in Figure 1.8(a). On the other hand, if a tangential stress such as shear due to friction is present, the
fluid deforms and continues moving. Such a flow has an oblique stress because the resultant
force from the tangential shear and normal pressure acts at an oblique angle.
3A
stress, by definition, is a force per unit area.
1.4. CONTINUUM APPROXIMATION
13
Figure 1.7: The flow of traffic can be modeled as a fluid whose density, defined as the number
of automobiles per unit distance, varies with speed. [Photographs courtesy of California
Department of Transportation]
For example, consider flow in a channel as shown in Figure 1.8(b). If we apply different
pressures p1 > p2 at the ends of the channel, frictional forces develop at the channel walls
(and, in fact, throughout the channel) to balance the pressure difference. As a result, the
initially rectangular section of fluid moves at different velocities across the channel, and thus
undergoes significant distortion. By contrast, a solid will resist a shearing force until it yields.
...
Resultant Stress .............
p
p
.......................
......................
......................
... .. ... ... ... ... .. ... ... ... ...
... ... ... ... ... ... ... ... ... ... ...
.. ... ... ... ... ... ... ... ... ... ...
.
. .
....... ........ ........ ........ ........ ........ ....... ........ ........ ....... ........
. . . . . . . . . . .
........................
........................
........................
........................
.......................
.......................
.......................
........................
p2 + τ
....... ....... ....... ....... ....... ....... ....... ....... ....... ....... .......
.... ... ... ... ... ... ... ... ... ... ...
.. ... ... ... ... ... ... ... ... ... ...
... ... ... ... ... ... ... ... ... ... ...
.
..
..........................
..
.............................
...
..............................
.
..........................
.
..
$
p
p1
............................................
........................
p
(a) Normal Stress
%
.
... ..
2 .... .....
.. .
.. ....
.
..... .....
..........................
p
τ
τ
.........................
..... ....
... ..
... ...
... ..
... ...
... ..
... ...
.....
2
.....
.
Resultant Stress
.
..
........................
..
.............................
...
..............................
.
........................
...
..
$
........................
p2
%
p
p2 + τ
(b) Oblique Stress
Figure 1.8: Fluid response to normal and oblique stresses. The forces are not drawn to scale
in (b)—shear stresses are generally much smaller than pressure. Also, in (b), the solid lines
indicate the boundaries of the moving fluid that was initially contained within the volume
denoted by the dashed lines.
1.4 Continuum Approximation
In reality, a fluid is made up of molecules. The distance between molecules in a gas is
very large compared to the diameter of a molecule. For a liquid, the molecular spacing is
comparable to molecular diameter. The molecules are not fixed in space. Rather, they are in
constant motion, colliding with one another. In a given volume, we thus have a collection of
molecules with empty space between them. This stands in distinct contrast to a continuous
14
CHAPTER 1. INTRODUCTION
substance in which the fluid mass density (mass per unit volume) would vary smoothly from
one point to another.
Since a fluid is not a continuous substance, in principle, we must apply the appropriate
laws of motion to each molecule to describe the overall motion. This would be an extremely
difficult task given the large number of molecules present for any practical flow. Table 1.5
quantifies just how large a number this would be for water and air. Note that, by definition, the
number density, n, is the number of molecules per unit volume. The values quoted correspond
to Standard Temperature and Pressure (STP). By definition, STP is 0o C and 1 atm. One
way of computing the motion of such a large system of particles is to use the mathematically
complex theory of statistical mechanics.
Table 1.5: Molecular Number Density for Water and Air
Fluid
Water
Air
Number of Molecules
3.34 · 1019
2.69 · 1019
Volume
1 mm3
1 cm3
Number density, n
3.34 · 1019 mm−3
2.69 · 1019 cm−3
To simplify our task, we introduce the continuum approximation, a concept that finds
its origins with Swiss mathematician Leonhard Euler (1707-1783). The continuum approximation is used in many engineering disciplines such as solid mechanics, thermodynamics and
electrodynamics. Using this concept means we regard a fluid as being a continuous substance
as opposed to being made up of discrete molecules. That is, we regard the fluid as being an
infinitely divisible substance with no empty space between particles. We pretend that the basic
properties (e.g., pressure, density, viscosity) of the smallest subdivisions of a fluid approach a
unique limiting value as we reduce the size of the subdivision. In other words, we assume a
definite value of each fluid property exists at each point in space.
Our justification for making this approximation is as follows. We can average random
thermal motion of molecules provided the number of molecules in a fluid particle is very
large, where we define a fluid particle as a volume of fluid whose size is extremely small
compared to the characteristic dimension of the flow under consideration.
∆m
∆V
.....
.........
...
..
..
.....
......
.......
......
....
........
.......
........
......
....
........
.
.
.
.
.
.
.... ..... .. ........ ......
.
.
.
...........
.. .. ..... .. ... ......
...............
... ......... ... ... .. ... ................................................................................................................................................................................................................................................................
... ....... ....... ... .... ....
.... . ......... .........
... ....
..
....
...
...
..
...
.....
..
...
..
..
...
............................
0
∆V
∆Vu
∆V
Figure 1.9: Continuum definition of mass density at a point in a fluid.
Figure 1.9 illustrates the limiting process we use to define a fluid property such as mass
density, ρ, where mass density is mass per unit volume at each point in a given fluid. In
terms of a fluid particle, if its mass is ∆m and its volume is ∆V , there is a limiting volume,
1.5. MICROSCOPIC AND MACROSCOPIC VIEWS
15
∆V , below which an insufficient number of molecules can be found to define a meaningful
statistical average. Also, as shown in the figure, for larger values of ∆V , say ∆V > ∆Vu ,
the ratio of ∆m to ∆V will vary if the fluid has nontrivial spatial variations in its properties.
In the continuum limit, we postulate that a wide separation of scales exists, i.e., that
∆V
∆Vu
(1.13)
and we define the mass density by
∆m
(1.14)
∆V →∆Vf ∆V
When there is a wide separation of scales, the limiting process defined in Equation (1.14) is
well defined, with ∆m/∆V approaching a constant limiting value at each point in the fluid.
We have a continuum provided, of course, that we can define a fluid particle in which a
sufficient number of molecules exists to define statistical averages of molecular motion. The
vast majority of practical fluid-flow applications satisfy this constraint.
ρ=
lim
Example 1.4 Consider flow past a sphere of 1 cm diameter in a standard wind tunnel. Determine
the number of molecules in a volume the size of a dust particle (diameter 1 micron = 10−4 cm)
for air at STP. If we regard this small of a volume as a typical “fluid particle” for analyzing flow
past the sphere, does the continuum approximation apply?
Solution. Since the volume of the dust particle is 16 π · 10−12 cm3 and Table 1.5 tells us there are
2.69 · 1019 molecules in 1 cm3 of air at STP, the number of molecules is
N=
1
molecules
π · 10−12 cm3 × 2.69 · 1019
= 1.4 · 107 molecules
6
cm3
Clearly, we can define meaningful statistical averages if we have such a large number of molecules
in our sample. Thus, the continuum approximation is valid for this flow.
Example 1.5 Now let’s move the sphere 200 miles above the Earth’s surface. At this altitude,
there are approximately 2 · 104 molecules in each cm3 . Again, compute the number of molecules
in a volume the size of a dust particle (diameter 1 micron = 10−4 cm). Does the continuum limit
apply in this case for analyzing flow past the sphere?
Solution. Since the volume of the dust particle is 16 π · 10−12 cm3 and there are 2 · 104 molecules
in 1 cm3 , the number of molecules is
N=
1
molecules
= 1.0 · 10−8 molecule
π · 10−12 cm3 × 2 · 104
6
cm3
At this altitude, our dust-particle sized volume contains 10−8 molecule, i.e., to find a single molecule
we need a volume a hundred million times larger than that of a dust particle. The continuum limit
is meaningless for this flow.
1.5 Microscopic and Macroscopic Views
With the continuum approximation, we view the fluid from the macroscopic level. By contrast,
in viewing the fluid on a scale of the order of a molecule, we observe things on the microscopic level. Hence, we refer to continuum fluid properties like mass density as macroscopic
properties. In this section, we will explore the manner in which density, pressure, temperature
and other thermodynamic variables relate to what actually happens at the microscopic level.
16
CHAPTER 1. INTRODUCTION
1.5.1 Density
Mass density, usually called density, is denoted by ρ and has dimensions of mass per unit
volume. Thus,
ρ = mn
(1.15)
where m is mass of a molecule and n is the number density, i.e., the number of molecules
per unit volume (see Table 1.5). Values of ρ for air and water at atmospheric pressure and a
temperature of 20o C (68o F) are as follows.
3
ρ=
3
1.20 kg/m (0.00234 slug/ft ), Air
3
3
Water
998 kg/m (1.94 slug/ft ),
(1.16)
The density of a liquid is a weak function of pressure and depends mainly on temperature.
Table A.2 in Appendix A includes densities of common liquids, while Table A.3 includes the
variation of density with temperature for water.
The density of a gas varies with both pressure and temperature. The perfect-gas law,
which is given below as Equation (1.25) and in Appendix A as Equation (A.1), can be used
to compute the density of a given gas.
There are three additional quantities related to density that are used in fluid-flow applications, viz.,
• Specific volume, υ: Used primarily in thermodynamics, this quantity is the volume per
unit mass. Thus,
1
υ=
(1.17)
ρ
• Specific weight, SW : Used primarily for flows with constant density, this quantity is
weight per unit volume. Thus, denoting gravitational acceleration by g,
(1.18)
SW = ρg
• Specific gravity, SG: This is a dimensionless representation of density. For a liquid,
it is the ratio of fluid density to the density of water at 4o C. For a gas, it is the ratio of
gas density to the density of air at 20o C and 1 atm. Hence, we have
SG =
ρ
ρref
3
where ρref =
3
1000 kg/m (1.94 slug/ft ),
Liquid
3
3
1.20 kg/m (0.00234 slug/ft ), Gas
(1.19)
Example 1.6 Using SI units, compute the specific volume, specific weight and specific gravity for
glycerin.
Solution. From Table A.2 the density of glycerin is 1260 kg/m3 and Table 1.4 tells us the gravitational acceleration, g, is 9.807 m/sec2 . Thus, noting that 1 kN = 1000 kg·m/sec2 in computing
specific weight, we have
3
1
−4 m
υ=
3 = 7.94 · 10
kg
1260 kg/m
kg
m
kN
SW = 1260 3 · 9.807
= 12.36 3
m
sec2
m
1260 kg/m3
SG =
= 1.26
1000 kg/m3
1.5. MICROSCOPIC AND MACROSCOPIC VIEWS
17
Example 1.7 Using USCS units, compute the specific volume, specific weight and specific gravity
for a gas whose density is 0.00125 slug/ft3 .
Solution. Table 1.4 tells us that g = 32.174 ft/sec2 . Noting that 1 lb = 1 slug·ft/sec2 in computing
specific weight, there follows
υ=
1
ft3
3 = 800
slug
0.00125 slug/ft
ft
lb
slug
· 32.174
= 0.0402 3
sec2
ft3
ft
0.00125 slug/ft3
SG =
= 0.534
0.00234 slug/ft3
SW = 0.00125
1.5.2 Temperature
In describing the motion of a molecule, we separate its instantaneous velocity, u, into two
parts, i.e., u = U + u . The first is the bulk-motion or mean velocity, U, and second is the
fluctuating velocity, u . The latter contribution represents random molecular fluctuations. By
definition,
U =<u>
(1.20)
where < u > denotes statistical average of u. The kinetic-theory definition of temperature, T ,
[cf. Jeans (1962)] is
3
kT = < 12 mu · u >
(1.21)
2
where k is Boltzmann’s constant and m is molecular mass. That is, the temperature of a
fluid is directly proportional to the average kinetic energy of the molecular motion. Standard
temperature is the value at which water freezes, viz.,
Tstandard = 0o Centigrade = 32o Fahrenheit
[T (o F) = 1.8T (o C) + 32o ]
(1.22)
[T (o R) = 1.8T (K)]
(1.23)
In terms of absolute temperature, we have
Tstandard = 273.15 Kelvins = 491.67o Rankine
1.5.3 Pressure
For a fluid at rest we observe that the fluid exerts a pressure on the walls of its container. In
fact, the only force exerted by the fluid is everywhere normal to the surface of the container.
We also observe that if the fluid is moving, it exerts an oblique force on the walls of the
container. We define the pressure, p, as the normal force per unit area exerted by a given fluid
particle on its immediate neighbors, whether or not the fluid is moving.
Pressure is a scalar quantity and thus acts equally in all directions at a given point in
the fluid. It can change from point to point in a given flow as required to balance local
accelerations. The force attending pressure is formed as the product of the pressure, the area
of the surface on which it acts and the unit normal to the surface. We refer to pressure as
a normal stress. We will take up the notion of oblique stresses when we discuss effects of
friction in Section 1.9.
18
CHAPTER 1. INTRODUCTION
........
.......
...........
..............
v
f
p
mu
.
.......
.......
...........
.............
....
....
....
....
....
....
....
....
..
... .
... .
.
....
.
.
.
..
.
.
.
...
.... .
....
.......
... ..
¦¦EE
¦¦EE
¦E
¦¦EE ¦E
¦
E
¦
E
¤
C
¦E ¦ E ¦E ¦ E ¦ E ¦E ...............
v
f
time
Figure 1.10: Impulses from molecules striking a container wall in the limiting case of number
density n
1.
At the microscopic level, fluid molecules are continuously moving in a random manner.
Collisions with a container wall give rise to an instantaneous force (impulse) on the wall. As
illustrated in Figure 1.10, if the number density is small, the resulting “pressure” appears as
a series of spikes. The magnitude of the impulse increases as the angle of the molecule’s
trajectory approaches a right angle, which involves the maximum momentum change normal
to the wall.
As depicted in Figure 1.11, for a very large number of collisions the effect of the collisions
is similar to a continuous push or pressure on the wall. This is the continuum limit (n
1),
and we use a statistical average, < p >, to define the pressure. Pressure is expressed in many
different units. For example, the standard value of pressure in the atmosphere, pa , is
⎧
1
⎪
⎪
⎪
⎪
760
⎪
⎪
⎨
14.7
pa =
101
⎪
⎪
⎪
⎪
760
⎪
⎪
⎩
2116.8
atm
torr
psi
kPa
mmHg
psf
(atmosphere)
(Torr)
(pound per square inch)
(kiloPascal)
(millimeter of mercury)
(pound per square foot)
(1.24)
where one Pascal = one Newton per square meter. Pressure is often quoted in terms of
absolute (psia) and gage (psig) values, with the latter relative to the atmospheric pressure.
Thus, p = 14.9 psi is an absolute pressure of 14.9 psia and a gage pressure of 0.2 psig.
p
<p>
.
......
.........
.
.
..
.. ......
.
.
....
...... .......
....
...... ...... ......
...............
.....
............... ....
....
.
.. ..... ... ..... .... ...
.....
.
............ .... ....
.... .... ........ .... .......
... .... .......... .... .......... .........
....
...... . ....... ...... ................ .... .... ...............
.. .... .. .. .... .....
. .... .. .. ..... ........... .. ..... .............. .......
...
.
.....
........... .. ...
.
.
.
.
.
............ ................ .. ........................... .. .. ....... ............... .. ........... .. .. .. .. ....... ...... .............. ....... .. .. ...... ...... .. ...... ......... ......... ...... ....... .. .... .. .. .. ............ ............... .. .. ...... .............................
.. ..
.... .... ..
........ ... ...... .... ...
..........
...... .
..... ... .... ..... ...... .....
.... ....... ................ ....... ................... ................. .....
.......
... ...
................
.... ....
.
.....
...
.
. . .. . ...... ..
.....
...
.
.....
......
.....
..
.
....
...........
..
.....
..................
time
Figure 1.11: Pressure from molecules striking a container wall in the limiting case of number
density n
1. p is the instantaneous pressure and < p > is the statistical average.
1.5. MICROSCOPIC AND MACROSCOPIC VIEWS
19
1.5.4 Vapor Pressure
There is a special value of the pressure known as the vapor pressure, pv , that manifests itself in
the familiar phenomena of evaporation and boiling. It also affects a more subtle phenomenon
known as cavitation.
To understand the concept of vapor pressure, consider the following. Imagine that we pour
a liquid into a closed container and measure the pressure in the space above the liquid. As
time passes, we will observe an increase in the pressure. At some point the pressure will reach
a constant value. By definition, this is the vapor pressure for the liquid.
In our hypothetical closed container, some of the molecules within the liquid have sufficient
kinetic energy to overcome the attractive forces between the molecules. As the number of
molecules that have escaped increases, the number of collisions of the vapor molecules with
the liquid surface increases and some of the vapor molecules rejoin the main body of liquid.
Equilibrium is achieved when the number of molecules rejoining the liquid equals the number
leaving to enter the vapor phase.
Clearly, vapor pressure is dependent upon the strength of the intermolecular attractive
forces. These forces are very strong for liquids like mercury and glycerin. Consequently,
a relatively small number of molecules enter the vapor phase and pv is very small (see Table A.11 in Appendix A). By contrast, the intermolecular forces are much weaker for carbon
tetrachloride and gasoline, resulting in much larger pv .
Since the molecular kinetic energy is greater at higher temperature, more molecules can
escape, wherefore vapor pressure is correspondingly greater. And indeed, we observe that
vapor pressure increases with temperature. Table A.11 in Appendix A tabulates values of
pv for water. Focusing on the pressure within a given liquid, p, we have three different
possibilities that depend upon how it compares to pv .
• Evaporation (p > pv ): If a liquid rests in an open container, p > pv and fluid will
continually enter the vapor phase. Unlike the situation in a closed container, most of the
vapor above the liquid escapes to the atmosphere. There is a one way transition from
liquid to vapor phase, and the liquid evaporates.
• Boiling (p ≤ pv ): When we increase the temperature of a liquid, vapor bubbles begin
to form within the liquid. Since the pressure in an interior bubble is equal to the vapor
pressure, such a bubble will rapidly collapse if the pressure in the fluid exceeds pv .
On the other hand, when the pressure in the fluid is such that p ≤ pv , the bubble
will remain or expand and the liquid boils. The temperature at which pv equals the
pressure of the atmosphere is the boiling point. Water boils at 212o F at sea level
where atmospheric pressure is 14.7 psi. At altitudes far above sea level, air pressure
drops appreciably with an attendant drop in boiling-point temperature. For example, at
the top of Alaska’s 17000-foot Mount McKinley, atmospheric pressure is approximately
7.6 psi, and Table A.11 shows that water will boil at 180o F. Since cooking occurs at a
lower temperature, it would take longer to cook a hard-boiled egg, or any food requiring
boiling, at the top of Mount McKinley than it would in your kitchen.
• Cavitation (p ≤ pv ): When a fluid is flowing, boiling can be caused at any temperature
by regions of very low pressure. This occurs, for example, in valves, pumps and marine
propellers. For flow through or past such devices, it is possible to have the local pressure
drop below the vapor pressure. This can cause severe problems because, when the vapor
bubbles are swept into higher-pressure regions in the flow, they will suddenly implode.
If the implosion occurs close to a solid surface, it can—over a period of time—cause
20
CHAPTER 1. INTRODUCTION
structural damage. Figure 1.12 shows cavitation emanating from the tips of a marine
propeller. The bubbles are shed from the low-pressure regions near the tips of the
propeller blades and form a helical pattern as the fluid moves from left to right.
Figure 1.12: Tip cavitation from a marine propeller. [Photograph courtesy of Garfield Thomas
Water Tunnel, Pennsylvania State University]
Example 1.8 A researcher has designed a test rig to demonstrate the effect of temperature on
cavitation. In the test section of the test rig, the pressure varies as
p = pa 1 −
7
8
s
c
2
where p is test-section pressure, pa is atmospheric pressure, s is distance from the beginning of
the test section and c is test-section length. The fluid used in the experiment is water.
(a) Verify that there will be no cavitation when the temperature in the rig is T = 50o C.
(b) If temperature increases to 60o C, compute the value of s/c at which cavitation occurs.
Solution.
(a) Clearly the pressure decreases as s/c increases and assumes its minimum value when s/c = 1.
Thus, since atmospheric pressure is pa = 101 kPa,
pmin = (101 kPa) 1 −
7
= 12.625 kPa
8
Reference to Table A.11 tells us that the vapor pressure of water at 50o C is pv = 12.29 kPa. Since
the pressure never drops below pv , there will be no cavitation.
(b) When the temperature increases to 60o C, Table A.11 tells us that the vapor pressure increases
to pv = 19.84 kPa. Thus,
(19.84 kPa) = (101 kPa) 1 −
7
8
s
c
2
Solving for s/c yields
s
=
c
8
7
101 kPa − 19.84 kPa
101 kPa
= 0.958
Therefore, cavitation occurs at a point 95.8% of the way through the test section.
1.6. THERMODYNAMIC PROPERTIES OF GASES
21
1.6 Thermodynamic Properties of Gases
When our fluid is a gas, there are some additional useful thermodynamic relations and properties of interest in the study of fluid mechanics. For most applications, we treat the gas as
though it is a perfect gas. Also, when we are interested in heat transfer, there are several
energy-related properties of interest to us.
If the intermolecular forces in a gas are negligible and intermolecular collisions are perfectly
elastic, the pressure, density and temperature are related by the perfect-gas equation of state,
(1.25)
p = ρRT
where R is the perfect-gas constant. For air, the value of R is
Rair = 287 J/(kg·K) [1716 ft·lb/(slug ·o R)]
(1.26)
Table A.1 in Appendix A includes values of R for common gases.
Alternatively, the perfect-gas law can be stated in terms of the universal gas constant, R,
according to
ρ RT
(1.27)
p=
M
where M is the molecular weight of the gas. Note that the units of molecular weight are
kg/(kg-mole) [slug/(slug-mole)]. The value of the universal gas constant is
R = 8314
J
ft · lb
= 49700
kg-mole · K
slug-mole ·o R
(1.28)
Example 1.9 The pressure in a one cubic foot tank of compressed helium is 45 psi. The temperature
inside the tank is 77o F. Compute the density of the helium. Also, compute the weight of the helium
in the tank, expressing your answer in ounces.
Solution. We use the perfect-gas law to compute the density so that
ρ=
p
RT
Reference to Table A.1 tells us that R = 12419 ft·lb/(slug·o R). Also, converting to absolute
temperature, we have T = (77 + 459.67)o R = 536.67o R. Then,
45
ρ=
lb
in2
144
ft · lb
12419
slug·o R
in2
ft2
(536.67o R)
= 9.7226 · 10−4
slug
ft3
Letting V denote the tank volume, the weight, W of the helium within the tank is
W = ρgV = 9.7226 · 10−4
slug
ft3
32.174
ft
sec2
1 ft3
16
oz
lb
= 0.50 oz
22
CHAPTER 1. INTRODUCTION
We will make use of other thermodynamic quantities and notation. For example, the heat
contained by a fluid is quantified in terms of the specific internal energy, e. The often omitted
prefix “specific” means “per unit mass.” In general, the basic postulates of thermodynamics
tell us that e is a function of any two independent thermodynamic state variables, a point we
will address in Chapter 7. The internal energy is usually written as a function of temperature
and specific volume, υ = 1/ρ, so that, in general, e = e(T, υ).
There is a second thermodynamic quantity that characterizes the heat contained by a fluid.
It is called specific enthalpy and is denoted by h. By definition,
(1.29)
h = e + p/ρ
In contrast to internal energy, h is usually written as a function of temperature and pressure,
i.e., h = h(T, p). Enthalpy and internal energy have units of J/kg (ft·lb/slug).
There are two specific-heat coefficients denoted by cv and cp . By definition,
cv =
∂e
∂T
and
cp =
υ
∂h
∂T
(1.30)
p
where cv is appropriate for a constant-volume process and cp for a constant-pressure process.
Both of the specific-heat coefficients have units of J/(kg·K) [ft·lb/(slug·o R)] and, in general,
depend upon T .
Finally, the ratio of the specific-heat coefficients is denoted by γ (the symbol introduced
by Rankine) and referred to as the specific-heat ratio. Thus, we write
γ=
cp
cv
(1.31)
There are two limiting cases that are of interest regarding internal energy, enthalpy, the
specific-heat coefficients and the specific-heat ratio.
• Thermally-perfect gas: For a wide range of practical conditions, we find that e = e(T )
and h = h(T ) for many gases. It is true, for example, for air at temperatures up to
about 2000 K (3600o R). A gas is said to be thermally perfect when e and h depend
only upon T .
• Calorically-perfect gas: Except at very high temperatures, cv and cp are constant for
many gases. When cv and cp are constant, we say the gas is calorically perfect. It
follows that, for a calorically-perfect gas, e = cv T , h = cp T and γ = constant.
Table A.1 in Appendix A includes values of γ and cp for common gases.
1.7 Compressibility
If we hold temperature constant and apply a pressure to a container filled with gas, its density
will change according to
p
ρ=
(1.32)
RT
so that the density change, ∆ρ, is given by
∆ρ =
∆p
RT
(constant T )
(1.33)
1.7. COMPRESSIBILITY
23
By increasing p, we compress the gas. The formal definition of the compressibility, τ , is
τ=
1 ∂ρ
ρ ∂p
(1.34)
Some authors prefer to work with the bulk modulus, Ev , which is the reciprocal of the
compressibility. For our isothermal perfect gas, we have ∂ρ/∂p = 1/(RT ) so that
τ=
1 1
1
·
=
ρ RT
p
(1.35)
Compressibility is a property of the fluid. The compressibility for gases is much larger than
corresponding values for liquids. At atmospheric pressure and constant temperature,
τ=
1.00 · 10−5
4.65 · 10−10
m2 /N (4.79 · 10−4 ft2 /lb), Air
m2 /N (2.23 · 10−8 ft2 /lb), Water
(1.36)
Table A.4 in Appendix A lists τ for several common fluids.
Example 1.10 Estimate how much greater a pressure must be applied to water, relative to that
required for air, to obtain a small fractional change in ρ.
Solution. Since Equation (1.34) implies ∆ρ/ρ = τ ∆p, we have
(∆ρ/ρ)H2 O
(∆ρ/ρ)air
=
τH2 O ∆pH2 O
τair ∆pair
=⇒
∆pH2 O
τair (∆ρ/ρ)H2 O
=
∆pair
τH2 O (∆ρ/ρ)air
So, to achieve the same fractional change in density, ∆ρ/ρ, the pressure change varies inversely
with compressibility. Appealing to Equation (1.36), we find
∆pH2 O
4.79 · 10−4 ft2 /lb
=
= 2.15 · 104
∆pair
2.23 · 10−8 ft2 /lb
Fluids that require very large changes in pressure to cause even a minor change in density
are termed incompressible. In a mathematical sense, we can say incompressible flow occurs
when τ → 0, so that ∆ρ = ρτ ∆p → 0. Hence, for our immediate purposes, we will consider
an incompressible flow to be constant-density flow. As we will see in Chapter 8, this is not
the conventional definition. It would imply that motion in a “stratified” medium such as the
ocean, whose density varies with depth, should be classified “compressible.” This is incorrect.
When we study compressibility effects we will find that a better definition of incompressible flow is very low Mach number flow, where Mach number is the ratio of flow speed to
the speed of sound. This is a far better indicator of incompressibility, and is satisfactory for
most practical applications.4
The Mach number appears in a more precise differentiation between incompressible and
compressible flows because it tells us something about how the flow adjusts to changes. If
there is a disturbance in the flow caused, for example, by the motion of an object, waves
propagate through the flow. This mechanism permits the flow to adjust itself and reach a
new steady equilibrium state corresponding to the altered flow conditions. For a compressible
fluid, these waves travel at a finite speed. In the case that the disturbance is weak, the wave
travels at the speed of sound.
4 Even this definition has its limitations. For example, flow in an internal-combustion engine occurs at low Mach
number, yet has large density changes caused by the very high pressure within the combustion chamber.
24
CHAPTER 1. INTRODUCTION
As we will show in Chapter 8, we can calculate the speed of sound, a, for any perfect gas
from the following equation.
a=
(Perfect gas)
γRT
(1.37)
where γ is specific-heat ratio defined in Equation (1.31), R is the perfect-gas constant (see
Table A.1) and T is absolute temperature. Now, using the perfect-gas law and Equation (1.35)
for compressibility, τ , the speed of sound in a gas thus depends upon τ according to
γ
ρτ
a=
(1.38)
So, for a gas, the speed of sound approaches ∞ as τ → 0. Although we must rely upon
measurements, the same proves to be true for liquids. Table A.4 in Appendix A lists the speed
of sound for several common fluids and shows, for example, that the speed of sound in water
is more than four times the speed in air. Thus, the speed of sound for an incompressible fluid
is large, so that acoustic waves travel at very high speed, and the entire flow will approach its
new equilibrium state very quickly.
Until we specifically address effects of compressibility in Chapter 8, most of our applications will be for incompressible flows. This greatly simplifies the equations we have to deal
with and actually precludes the need to include the energy principle for most problems. This
means that incompressible-flow problems can be solved using only the mass and momentum
principles. This by no means limits our study to a small class of fluid-flow problems. Incompressible flow covers the flow of liquids and “low-speed” gas flows, where low speed means
the Mach number is less than about 0.3.
Example 1.11 Using the sound speeds listed in Table A.4, compute the Mach numbers for a torpedo
with a speed of 40 mph, an automobile traveling at 65 mph, a Cessna 310 flying at 200 mph and
a bullet traveling through air at 1500 mph.
Solution. Table A.4 tells us that the speed of sound for water is aH2 O = 4859 ft/sec and for air
it is aAir = 1119 ft/sec. First we must convert speed of sound from ft/sec to mph. Reference to
Table 1.4 shows that 1 mph = 1.467 ft/sec. Hence,
aH2 O
=
(4859 ft/sec)
1 mph
1.467 ft/sec
= 3312 mph
aAir
=
(1119 ft/sec)
1 mph
1.467 ft/sec
= 763 mph
The Mach number, M , is the ratio of the object’s speed to the speed of sound. Therefore, for the
four objects, the Mach numbers are as follows.
⎧
(40 mph)/(3312 mph)
⎪
⎨
(65 mph)/(763 mph)
M=
(200
mph)/(763 mph)
⎪
⎩
(1500 mph)/(763 mph)
=
=
=
=
0.012,
0.085,
0.262,
1.966,
Torpedo
Automobile
Cessna 310
Bullet
For the torpedo, automobile and Cessna 310, the Mach number is less than 0.3. Consequently,
fluid motion about these vehicles can be considered to be incompressible. By contrast, the bullet is
traveling close to Mach 2 and compressibility effects cannot be ignored.
1.8. SURFACE TENSION
25
1.8 Surface Tension
There is an interesting effect that occurs near liquid-gas and immiscible-liquid interfaces, viz.,
surface tension. At such an interface, we find the surface to be like a stretched membrane. An
insect can walk on water (see Figure 1.13), a steel needle placed gently in a pan of water floats,
water “beads up” on your freshly waxed car. These are all commonly observed examples of
surface tension.
Figure 1.13: A water strider uses surface tension to support its weight. The photo (left) and
sketch (right) show the surface distortion where the strider’s legs contact the water surface.
Surface tension is actually a very complex phenomenon, involving concepts from physical
chemistry [cf. Adamson (1960)]. A simplified explanation that captures the dominant physics
is as follows. On the one hand, as indicated schematically in Figure 1.14, molecules within
the main body of the fluid are surrounded by molecules with attractive forces5 equal in all
directions. On the other hand, molecules at the surface are attracted only by molecules from
within the fluid and thus exhibit a stronger attraction for other surface molecules to achieve an
overall balance of molecular forces. This modified “force field” results in the surface behaving
like a stretched membrane.
Fluid 1
z.......................... z.......................... z.......................... z.......................... z.......................... z.......................... z
Fluid 2
z
z
z
......
......
.....
........
...........................
......
......
.....
.......
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
.
........
...
...
......
........ .........
. ......
..........................
. ......
........ ..........
.......
....
..
.......
.
Figure 1.14: Intermolecular forces are skewed near the interface between two different fluids
and cause the interface to behave like a stretched membrane.
We can thus describe surface tension mathematically by making an analogy to a membrane,
in which a uniform tensile force, σ, acts tangent to the surface. However, there is no physical
membrane present—only a distribution of molecular forces that acts like the tension in a
membrane. Note that, by definition, σ has dimensions of force per unit length.
5 These forces, known as Van der Waals forces, are electrical in nature. They are caused by the asymmetrical
charge distribution that exists within molecules.
26
CHAPTER 1. INTRODUCTION
For example, any curvilinear surface can be described in terms of two principal radii
of curvature, R1 and R2 (Figure 1.15). In analogy to a stretched membrane, the pressure
difference, ∆p, that can be supported by a surface tension, σ, on a general surface is given
by Laplace’s formula [see Landau and Lifshitz (1966)], which is
σ
σ
+
= ∆p
R1
R2
..
....... ......
..... ........ .........
......
.....................................................................
..................
..
.
.
.
.......
.
.
.................
..
..
................
.
.
.
....................
.
.
......
...
..
......................
....................
.
.
.
...
.......................................
................ ..
............
.
.
.
.
.
.
....
.
.
.......... .....
... ................................. ..
...
....
. ..
....
..
.... ........ .. ..
.. .. ...
...
...
.
.
...
...
...
...
... ...
...
...
... ...
...
..
2
... .
...
..
.
... ..
...
...
.
1
.
.
.
.
.
..
... .
.
... ...
......
(1.39)
σ
σ
σ
σ
σ
Cylinder: R1 → ∞,
∆p
σ
R
R
R
.....................................
............................. .... .......................................
....... ...... .... .... ...... ..... .
........ .. .... ... ... ... .. ...................
...........
.
.. . . . .. .
...
...
...
..
...
.
..
...
...
...
.
.
...
...
...
i
...
... ..
.
... ...
....
Sphere: R1 = R2
Figure 1.15: Stretched-membrane analogy for surface tension.
Alternatively, for a specific application, we can balance forces and arrive at the same
result with some added insight into the physics involved. The pressure within a droplet,
for example, is higher than the pressure outside. Surface tension supports the difference in
pressure. Consider the droplet depicted in Figure 1.16, with the pressure forces and surface
tension indicated. The net contribution from the atmosphere is pa times the projection of the
hemisphere onto a vertical plane. If this point is unclear, consider the limiting case ∆p = 0
and σ = 0 in which the atmospheric pressure on the two surfaces must be in balance. Thus,
F = (pa + ∆p) πR2 − pa πR2 − 2πRσ = 0
(1.40)
Simplifying, we find
∆pπR2 = 2πRσ
=⇒
∆p =
2σ
R
(1.41)
which is identical to what we obtained above using Laplace’s formula (with R1 = R2 = R).
...
..
.....................................................................................
..
... .........
...
.
...... .......
.
...
........
..
...
.....
.. ..
.
.
.
.
...
....
.........................
a
...
...
.
... . ........
...
...
.........
...
...
...
..
............................
...
...
.
.
...
..
.............................................................
...
..
..
.................
.
.
............................
.
.. .
.
...
.
.
.
...
.
.
...........................................................
..
....
..
.
.
.
.
.
.
.
.
.
.
..................... .
..
...
...
...........
...
...
... .......
..
...
...
..
..
.............................
.
.
.
.
...
...
...
...
.........
...... .........
...
...
.
.. .........
..
.........................................................................................
...
...
σ....
p + ∆p
σ
pa
σ
σ
Figure 1.16: Force diagram for a droplet.
1.8. SURFACE TENSION
27
Example 1.12 Compute the pressure within a 1-mm diameter spherical droplet of water relative
to the atmospheric pressure outside. The temperature is 20o C.
Solution. To apply Laplace’s formula, we make an imaginary cut through the center of the droplet
as shown in Figure 1.16. Since the principal radii of curvature are both equal to the sphere radius,
R = 5 · 10−4 m, and the surface tension for water is 0.073 N/m (Table A.5), Equation (1.41) gives
∆p =
2(0.073 N/m)
2σ
N
=
= 292 2 ≈ 0.003 atm
R
5 · 10−4 m
m
As a second interesting example of the effect of surface tension, consider the phenomenon
known as capillary action. A tube of perimeter P and cross-sectional area A is immersed
in a container filled with a liquid of density ρ as shown in Figure 1.17. The fluid is drawn
up into the tube a distance ∆h. Surface tension, σ, acts around the perimeter of the tube
and supports the weight of the column of fluid. The surface tension is tangent to the surface
as shown, making an angle φ with the tube surface. The vertical component of the surface
tension balances the weight of the column of fluid so that
P σ cos φ = ρg∆hA
=⇒
∆h =
Pσ
cos φ
ρgA
(1.42)
Very typically, the tube is a cylinder of diameter d, for which P = πd and A = πd2 /4. Then,
Equation (1.42) tells us
∆h =
4σ
cos φ
ρgd
(1.43)
(cylindrical tube)
The height is an average since our computation approximates the column of fluid as being
exactly rectangular. In reality, its shape is curved as shown in Figure 1.17, corresponding to
the commonly observed meniscus.
...
.
....
...........
...........
....
.
...
...
.
...
.
............
...
.
...
...
...
..
....
.....
.....
.
...
...
.................................... ................
........
....
...
...
..
...
φ
±
°
∆h
.
...............
...
..
.....
.................................................................................................................................
σ
d
...
...
...
...
.
.........
......
.
g
...
...
...
...
...
...
.
..
..... ........
..................................................................................................................................
...............
ρ
Figure 1.17: Capillary action—cross-sectional view. Note that σ has dimensions of force per
unit length. It acts along an arc, not on a surface.
In general, surface tension depends upon the surface material as well as the nature of the
two fluids involved, and it is a function of pressure and temperature. It is also considerably
affected by dirt on the surface. Table A.5 in Appendix A lists surface tension for several
common liquids. The angle φ is called the wetting angle or the contact angle. For water
interfacing air in a glass tube, we find that φ ≈ 0o . By contrast, for mercury interfacing air
28
CHAPTER 1. INTRODUCTION
in a glass tube, the surface tension acts downward with φ ≈ 129o . Similarly, for water, air
and paraffin wax, the wetting angle is 105o [cf. Sabersky et al. (1989)]. As a consequence of
having φ > 90o , the column of fluid is drawn downward by capillary action.
Example 1.13 Consider a cylindrical capillary tube made of glass with diameter d = 1 mm. The
liquid is water, and the tube is open to the atmosphere so that ρ = 998 kg/m3 , σ = 0.073 N/m and
φ = 0o . How many tube diameters will the water rise in the tube due to capillary action?
σ ..............
.....
.........
....
...
...
...
..
...
.
.
....
.....
...
..
...............................
−
σ
...............
....
...
...
...
..
ª
...
..
g .........
∆h
..
.........
.....
.............
...
.
...............................................................................................................
d
...
...
...
...
...
.
.
.
....................................................................................................................
............
ρ
Solution. Since φ = 0o , the surface-tension force is vertically upward as indicated in the figure.
Using Equation (1.43), the height of the column of water drawn up the tube is
(4)(0.073 N/m)(1)
= 0.0298 m
998 kg/m3 9.807 m/sec2 (0.001 m)
∆h =
Thus, we have
∆h
0.0298 m
=
≈ 30
d
0.001 m
Example 1.14 Now, change the fluid to mercury leaving all other conditions the same as in
Example 1.13. Using ρ = 13550 kg/m3 , σ = 0.466 N/m and φ = 129o , determine the change in
column height, ∆h, in terms of tube diameters.
.............
.................................................................................................................
..
¨
¥
ρ
d .............
.......................................................................................................................
...
..
..
¨
¥
|∆h|
.
.....
......
.
®
©
................................
......
.
.......
..... .
.....
.............
.
.
.
........
σ
........ .. 129
. .......
.....
.......
..............
.
o
...
.....
................
...
...
...
.........
......
.
g
σ
Solution. Since φ > 90o , the surface-tension force points downward as indicated in the figure.
Again using Equation (1.43), we obtain
∆h =
(4)(0.466 N/m)(−0.629)
= −0.0088 m
13550 kg/m3 9.807 m/sec2 (0.001 m)
Thus, we have
−0.0088 m
∆h
=
≈ −9
d
0.001 m
1.8. SURFACE TENSION
29
Example 1.15 A small U-shaped capillary tube with different diameters on each vertical segment
is inserted in a pan of water as shown in the figure. The fluid rises in the tube to its equilibrium
height of ∆h1 and ∆h2 in the larger and smaller tube segments, respectively. The pressure of
the air in the capillary tube is pb . Assume that the wetting angle, φ, is 0o , surface tension is σ
and water density is ρ. Also, the gravitational acceleration is g and atmospheric pressure is pa .
Determine the difference between ∆h2 and ∆h1 as a function of σ, ρ, g and d.
...............................................................................................................................................
............
........
.........
......
b
.
..
.........
......................................................................................................................................... ......
.
.
.
.
.
.
.
.
.
.
.
.
.
.................
............... ..............
............
...
................. .........
...
.... ...
.....
...
......... .................
..
..
.....
........
......
2
.
1
... . .
a
.
.
.
.
...
... .....
. .
.....................................................................
.......................................................................................................................... .....................................................................
p
2d
d
g
§¦
∆h
∆h
p
ρ
Solution. To solve, we must balance all of the forces acting on the fluid in each vertical
segment of the capillary tube. There are four forces acting, viz., the pressure force above
the column of fluid, the surface-tension force, the weight of the column of fluid and the
pressure force from below. Since the capillary tube to the left is cylindrical with diameter
2d, we have
−
pb πd2
+
Pressure above
−
2πdσ cos φ
ρg∆h1 πd2
Surface tension
Weight
pa πd2
+
=0
Pressure below
Thus, rearranging terms and noting that cos φ = 1 when φ = 0o , there follows
(ρg∆h1 + pb − pa ) πd2 = 2πdσ
ρg∆h1 + pb − pa =
=⇒
2σ
d
Therefore, the pressure difference, pb − pa , is
p b − pa =
2σ
− ρg∆h1
d
Similarly, for the tube to the right, which has diameter d, the force balance yields
−
pb πd2 /4
Pressure above
+
πdσ cos φ
−
Surface tension
ρg∆h2 πd2 /4
Weight
pa πd2 /4
+
Pressure below
Again, rearranging terms and using the fact that cos φ = 1, we have
(ρg∆h2 + pb − pa ) πd2 /4 = πdσ
=⇒
ρg∆h2 + pb − pa =
4σ
d
Therefore, the pressure difference, pb − pa , is
p b − pa =
4σ
− ρg∆h2
d
So, equating the two results above for pb − pa , there follows
4σ
2σ
− ρg∆h1 =
− ρg∆h2
d
d
=⇒
∆h2 − ∆h1 =
2σ
ρgd
=0
30
CHAPTER 1. INTRODUCTION
1.9 Viscosity
Fluids such as honey or molasses tend to flow very slowly down an inclined plane. Water
goes much faster. The former fluids are more viscous. Recall that for a container filled with a
fluid at rest we observe only normal stresses. If the fluid is moving, we also observe oblique
stresses on the walls of the container. In fact, these oblique stresses act everywhere in the
fluid if the fluid is moving.
To understand the origin of viscosity, we need to take another look at what goes on at the
molecular level. Referring to Figure 1.18, consider an imaginary plane in a fluid at y = 0,
which is shown as the shaded area. The figure shows one molecule above the plane (y > 0)
that moves, on the average, with velocity U1 . A second molecule lies below the plane (y < 0)
and it is moving with an average velocity U2 . We assume U1 > U2 .
As discussed in Section 1.5, in addition to their mean velocity, the molecules are moving
with random velocity fluctuations in both magnitude and direction. Of particular interest, this
includes fluctuations in the y direction that can carry molecules close to y = 0 across the
plane. Molecules migrating across y = 0 are typical of where they come from. That is,
molecules moving up bring a momentum deficit and vice versa. This incremental change in
momentum gives rise to a stress on a surface whose unit normal lies in the y direction and
that acts in the x direction. We define this type of stress as a shear stress.
y ........
z
. .
............... ..................
...
... .........
. .. ..
..
.........................................................
... ..... .. ...
..
................ ................
.... ....
u
t
...
.....
.....
.....
.....
.
.
.
.
.....
....
.....
. .
.....
............... ..................
.....
.
.
.
....
.
.. ........
.....
.....................................................
.....
.
.
.
.
.
.
.
.
.
.
.
.... ..... .. .......
....
.. .
.....
............... ...............
.....
.... ....
.....
.
.
.
.
.
.......
¡
¡
¡
¡
¡¡
t
u
U1
.....
.............
.....
.....
.....
....
.
.
.
.
.
......
...... ....
..... .......
......
.
.
.
.
....
.....
.....
.....
.....
.
.
.
.
..
x
U2
.....
............
Figure 1.18: Close-up view of molecular motion. Because of their random molecular motion,
molecules can migrate across the (shaded) plane, y = 0. The momentum transfer gives rise
to a shear stress.
When you step off of a conveyer belt (moving walkway), you experience this phenomenon.
You are typical of where you came from, namely a region where you had nonzero forward
momentum with respect to the stationary floor. As your feet touch the floor, you arrive in
a region where you have an excess amount of momentum and your body will lurch forward
until you regain your balance. Note that, assuming you step directly to your side to exit the
belt, the “force” on your body is normal to the direction you have chosen to move.
In a viscous fluid, the velocity in the x direction, u, varies continuously in the y direction.
For a majority of fluids encountered in engineering applications, we find that the shear stress,
τ , is proportional to the slope of the velocity profile, du/dy. Thus, we can say that
τ =μ
du
dy
(1.44)
where μ is the viscosity coefficient with units kg/(m·sec) [slug/(ft·sec)]. When τ is a linear
function of du/dy, we have what is referred to as a Newtonian fluid. For some fluids like
blood and long-chain polymers, we have a nonlinear relation between τ and du/dy. Such a
fluid is called a non-Newtonian fluid. A commonly used approximation for non-Newtonian
fluids is the Ostwald de Waele formula, which is
τ =K
du
dy
n
(1.45)
1.9. VISCOSITY
31
where K and n are empirically determined coefficients. Table A.10 of Appendix A lists K
and n for a variety of fluids. When n < 1, the fluid is referred to as pseudoplastic or shear
thinning. When n > 1, the fluid is called dilatant or shear thickening.
Often, especially for incompressible flows, we find ourselves dealing with the ratio μ/ρ.
This happens so frequently that we choose to dignify the ratio with its own symbol and name.
We define the kinematic viscosity, ν, [units m2 /sec (ft2 /sec)] as follows.
ν=
μ
ρ
(1.46)
Measurements indicate that viscosity of both liquids and gases varies with temperature but
shows virtually no dependence upon pressure. Usually dμ/dT and dν/dT are positive for a
gas while dμ/dT and dν/dT are negative for a liquid. Table A.6 in Appendix A includes
kinematic viscosities of common liquids and gases, while Table A.7 includes the variation of
ν with temperature for water.
There are numerous empirical formulas that are used in general engineering applications.
The following are three of the most commonly used equations. Note that, for all of these
formulas, T is absolute temperature, and is thus given either in Kelvins (K) or degrees Rankine (o R).
• Sutherland’s Law: This is an empirical equation for the viscosity of gases that is quite
accurate for a wide range of temperatures. The formula is
μ=
AT 3/2
T +S
(1.47)
where A and S are constants. The values of A and S for air in USCS units are
A = 2.27 · 10−8
slug
,
ft·sec·(o R)1/2
S = 198.6o R
(Air)
(1.48)
kg
,
m·sec·K1/2
S = 110.3 K
(Air)
(1.49)
The values for SI units are
A = 1.46 · 10−6
Table A.8 of Appendix A includes values of A and S for several common gases.
• Power-law viscosity: This is a common approximation for the viscosity of gases, which
assumes that viscosity varies with temperature according to
μ = μr
T
Tr
ω
(1.50)
where μr and Tr are reference viscosity and temperature, respectively. These constants
and the exponent ω depend on the gas. Table A.9 of Appendix A summarizes values
for several gases.
• Andrade’s equation: This is a useful empirical formula that is often used to approximate the viscosity of liquids. The viscosity varies with temperature according to
μ
= eTr /T
μr
(1.51)
As with the power-law formula for gases, μr and Tr are reference viscosity and temperature, respectively. Using measured values of μ, the reference values can be determined.
32
CHAPTER 1. INTRODUCTION
1.10 Examples of Viscosity Dominated Flows
There are some flows that are completely dominated by viscous effects. In this section,
we consider two such flows, i.e., two-dimensional Couette flow and axisymmetric HagenPoiseuille flow. Our discussion here is, of necessity, heuristic and limited to examining
properties of the solutions without benefit of a rigorous derivation. This is true because we
will not develop the viscous-flow equations of motion until Chapter 12. We examine the
rigorous derivation of the solutions for both of these flows in Chapter 13.
1.10.1 Couette Flow
Couette flow occurs when we have a fluid between two infinite parallel planes as shown in
Figure 1.19. The lower plane is at rest while the upper plane moves with constant velocity U ,
and the pressure is constant. This flow can be realized in a laboratory by having a conveyer
belt close to a stationary surface. If the distance h is very small compared to the length of
the belt, we approximate the idealized Couette-flow geometry provided we are not too close
to the ends of the belt.
¡
.........................................................................
...................................................
y
..
........
..
.....
..
.................................
..
.............................
¡
¡
¡
¡
¡
¡¡
.
.......
....
..
...
...
...
...
U
.................................................................
h.
...
...
...
...
...
...
.
.......
.
x
Figure 1.19: Couette flow—the lower wall is at rest and the upper wall is moving.
We expect the solution to depend only upon distance between the plates. If we look close to
the solid boundaries, we observe that the fluid “sticks” to the boundaries. This is known as the
no-slip surface boundary condition on the velocity. We will discuss this boundary condition
in more detail in Section 10.3. Thus, our solution must satisfy u(0) = 0 and u(h) = U ,
where u(y) is the velocity component in the x direction. Indeed, an exact solution to the
fluid mechanics equations of motion exists in which flow properties are functions only of y
(Section 13.1). The velocity varies linearly with distance across the channel, viz.,
y
u(y) = U
(1.52)
h
The corresponding shear stress is
μU
du
=
(1.53)
τ =μ
dy
h
Hence, the shear stress is not confined to the walls of the channel. Rather, it is constant across
the channel and acts throughout. We can rewrite Equation (1.53) in terms of nondimensional
parameters as follows.
τ
2
=
(1.54)
1
2
ρU
(ρ U h/μ)
2
The dimensionless grouping on the left-hand side of Equation (1.54) is called the skinfriction coefficient, and is usually denoted by cf . The dimensionless grouping in the denominator of the right-hand side of Equation (1.54) is called the Reynolds number, and is usually
1.10. EXAMPLES OF VISCOSITY DOMINATED FLOWS
33
denoted by Re. Experimental measurements are consistent with the linear-velocity-profile
solution provided
ρ Uh <
∼ 1500
(1.55)
μ
This is known as the laminar flow solution. At larger values of Reynolds number, this
solution is unstable to small disturbances, and the flow becomes turbulent. When the flow
is turbulent, the velocity and pressure vary rapidly with time and the effective shear stress is
much larger than the laminar-flow value given above. Chapter 14 includes an introduction to
turbulence.
The Couette-flow solution is often used as a “building-block” solution. Any application
that has flow through a small gap can be described in terms of the linear velocity variation in
the gap. Practical applications involving flow in a small gap include flow between a computer
hard-disk platter and read/write head, and the manufacturing process for a conventional recording tape. The Couette-flow solution (with the gap width, h, varying slowly in the streamwise
direction) also plays a role in the classical Lubrication Theory developed by Osborne Reynolds
in 1886.
Example 1.16 In manufacturing recording tape, the tape is coated with a lubricant. This is done
by pulling it through a narrow gap filled with the lubricant. Consider tape whose width out of the
page and thickness are w and h, respectively. The maximum tension force the tape can withstand
is T . The tape is centered in a gap of height 3h, the lubricant has viscosity μ, and the length of
tape within the gap is . What is the maximum velocity at which the tape can be pulled through
the gap?
.
.......
.......
.
.......
........
.
...............
....
.....
h
lubricant, μ
.......
........
.
tape
h
U
....................................................
h
...............................
..............................
Solution. The forces on the tape are the tension force, T , and the friction force on each
side of the tape. In steady state, these forces balance so that
T = 2τw w
where τw is the shear stress on the surface of the tape. Assuming the gap width is very
narrow so that h
, we can use the Couette-flow solution. Thus, the velocity in the
gap is u = U y/h, and the shear stress is
τw = μ
du
μU
=
dy
h
So, the tension force is given by
μU w
h
wherefore the maximum velocity at which the tape can be pulled through the gap is
T =2
U=
1 Th
2 μw
34
CHAPTER 1. INTRODUCTION
1.10.2 Hagen-Poiseuille Flow
Hagen-Poiseuille Flow is a second simple example of a flow completely dominated by viscosity. We consider flow through an infinitely long pipe with circular cross section (see
Figure 1.20). As with Couette flow, this subsection discusses properties of an exact solution
to the fluid-mechanics equations of motion (Section 13.3).
........
..... ....
..
...
..
...
..
..
.
..
..
...
..
....
..
....
.
..
..
...
...
...
...
...
.
.
...
.
..
...
..
...
..
...
...... .....
.......
p1
...
..... .....
..
..
..
..
...
....................................... ...
...
.
...
.
.
... ...
........
............................
............................
................................
L
τ
r
2r
...
..... ....... .....
........
..
..
...
...
.... ......................................
.
..
..
...
.
.
.
.
.
.
.
..
.
..... ................
................
..... . .....
.... ........ ..... ..
..
..........
.
...
.....
...
...
...
...
...
..
....
....
.............................................
....
...
..
...
....
.
.
..
...........
.
.
.
.
... ....
...
...
.....
...
................
p2
τ
2R
x
................................
Figure 1.20: Hagen-Poiseuille (pipe) flow.
Consider the flow in a cylindrical section within the pipe of radius r and length L as shown
in Figure 1.20. Assume the pressure on the left side of the section is p1 and the pressure on
the right side of the section is p2 . We choose our sign convention so that positive shear stress
on the cylindrical section is in the positive x direction as shown.
The pressure difference is balanced by the shear stress acting on the circumference of the
cylindrical section. Hence, we have
p1 πr2 − p2 πr2 + 2πrLτ = 0
(1.56)
Thus, the shear stress is given by
τ =−
(p1 − p2 ) r
L
2
(1.57)
So, when p1 > p2 , the shear stress is negative and therefore resists the motion. We can
determine the velocity profile, u(r), by integrating Equation (1.44) with r replacing y. The
integration is done subject to the no-slip boundary condition that tells us u(R) = 0. The
resulting expression for the velocity then becomes
u(r) =
(p1 − p2 ) 2
r2
R − r2 = um 1 − 2
4μL
R
(1.58)
where um is the maximum velocity, which occurs on the centerline (r = 0) and is given by
um = u(0) =
(p1 − p2 ) 2
R
4μL
(1.59)
We observe this (laminar-flow) solution experimentally provided the Reynolds number
based on um is as follows.
ρ um R <
∼ 2300
(1.60)
μ
If great care is taken to prevent transition to turbulence by keeping disturbances to a minimum,
laminar flow can be achieved for pipe flow at a Reynolds number as high as 40000 [Schlichting
and Gersten (2000)]. We will study pipe flow in greater detail in Chapters 7, 13 and 15.
1.11. FLUID-FLOW REGIMES
35
Example 1.17 The density and viscosity of human blood at body temperature are ρ = 1058 kg/m3
and μ = 3.3 · 10−3 kg/(m·sec), respectively. The maximum speed of blood in the aorta of a large
man is roughly um ≈ 0.30 m/sec. If the diameter of the aorta is d ≈ 3 cm, what is the Reynolds
number, ρum d/μ? Will the flow be laminar or turbulent?
Solution. For the given conditions, the Reynolds number based on diameter is
ρum d
=
μ
1058
kg
m3
0.30
3.3 · 10−3
m
(0.03 m)
sec
= 2885
kg
m · sec
To have laminar flow, the Reynolds number based on radius must be less than 2300, which corresponds to Reynolds number based on diameter of 4600. Therefore, the flow is laminar.
1.11 Fluid-Flow Regimes
Figure 1.21 presents an outline of the major regimes of classical continuum fluid mechanics. It
also serves as a “road map” for this text, indicating the chapters in which the various regimes
are discussed.
The first major subdivision of the discipline is between the flow of inviscid and viscous
fluids. An inviscid fluid is also referred to as an ideal fluid. We define an ideal fluid as one
in which the stresses acting are everywhere normal to the element of surface on which they
are measured, whether or not the fluid is moving. This is sometimes referred to as a perfect
fluid or a frictionless fluid. This is a drastic simplification as compared to a viscous fluid,
which is often referred to as a real fluid.
Continuum Fluid Mechanics
Chapters 1-15
Inviscid Flow
Viscous Flow
Chapters 5-9,11,15
Chapters 10,12-15
Laminar Flow
Turbulent Flow
Chapters 13-15
Chapters 14-15
Incompressible Flow
Compressible Flow
Chapters 5-7,9-14
Chapters 8,15
Figure 1.21: Major regimes of fluid mechanics.
36
CHAPTER 1. INTRODUCTION
For a moving fluid, the approximation of treating it as ideal cannot hold everywhere. For
fluids of small viscosity such as water and air, we observe two important facts. On the one
hand, since viscous stresses are proportional to velocity gradients [cf. Equation (1.44)], neglect
of viscous stresses breaks down where velocity gradients are large. On the other hand, the
flow around a streamlined body behaves almost as if the fluid were frictionless except in very
thin layers next to the surface and in a thin wake downstream. These near-surface layers were
discovered by Prandtl in 1904, who named them boundary layers. Velocity gradients are
very large in boundary layers. We will find out how thin boundary layers are in Chapter 14.
The concept of an ideal fluid is very useful, and much of our analysis in the first 11
chapters assumes effects of friction are unimportant. There are many interesting problems we
can solve ignoring effects of friction, and the attendant mathematical simplification permits
us to focus on the physical behavior of fluids without getting lost in the algebra.
However, there are problems for which an inviscid solution is completely unsatisfactory.
Inviscid flow past a sphere or cylinder, for example, bears little similarity to the flow observed
in nature. The inviscid solution predicts that there is no net force on the object, which is
contrary to physical observations. Even when we include viscosity, the flow past solid objects
falls into two different categories.
On the one hand, at very low flow speeds, fluid motion is very smooth, and flow patterns
remain unchanged from one instant to the next. This type of motion is termed laminar flow.
We have seen examples of exact laminar-flow solutions in Section 1.10, and several more are
presented in Chapter 13. On the other hand, at higher flow speeds, laminar flow becomes
unstable and undergoes transition to what is known as turbulent flow. Turbulence is a state
in which all flow properties vary rapidly with time, with a wide range of excited frequencies.
All laminar flows ultimately become turbulent, with “higher flow speeds” corresponding to
the mildest of motions such as a small puff of wind. Although extensive research efforts have
focused on the topic, to paraphrase Prof. Richard Feynman, turbulence remains one of the
most noteworthy unsolved scientific problems of modern times. Chapter 14 includes a brief
introduction to turbulence.
The final major subdivision addressed in this book is between compressible and incompressible flows. As discussed in Section 1.7, there are many incompressible-flow applications
in everyday life. It is equally true that compressible flow abounds in practical engineering
design and application. Transmission of gas in pipelines at high pressure, flow past a commercial airliner, and flow beneath an Indy 500 racer all involve compressible flow. Chapters 8
and 15 focus explicitly on compressible flow.
1.12 Brief History of Fluid Mechanics
The first part of this text focuses almost entirely on inviscid fluids. The second part turns to
the more complex analysis of viscous fluids. Approaching fluid mechanics in this manner, to
some extent, reflects the chronological development of the subject. The quantitative approach
to fluid mechanics began in earnest with Isaac Newton’s seventeenth-century formulation of
the laws of classical mechanics, including a postulate for how effects of friction should appear
in a fluid. Eighteenth-century mathematicians, most notably Bernoulli, d’Alembert, Euler,
Lagrange and Laplace, developed the theory of frictionless flow. Two unsettling results of
this early work were the Helmholtz Theorem and d’Alembert’s Paradox, proving that no
forces can develop for inviscid flow past an object in an unbounded fluid. This is, of course,
at variance with physical reality.
1.12. BRIEF HISTORY OF FLUID MECHANICS
37
Eighteenth- and nineteenth-century attempts at applying Newton’s law of friction to resolve
d’Alembert’s Paradox proved unsatisfactory, raising serious doubts about theoretical fluid
mechanics. Largely rejecting theoretical methods, the field of hydraulics was formulated by
researchers such as Chézy, Weber, Hagen, Poiseuille, Darcy and Weisbach. Their work was
based almost entirely on empiricism and experimentation, with little relation to basic physics.
It wasn’t until the end of the nineteenth century that a unified theory of fluid mechanics
including both theoretical and experimental methods took hold. The first important developments that helped unify the theory (with key contributions by Froude, Rayleigh and Reynolds)
were dimensional analysis and dynamic similitude. These techniques, although adding minimal physical understanding, helped correlate measurements and establish parallels between
experimental observations and the differential equations of motion.
Prandtl’s discovery of the boundary layer in 1904 resolved d’Alembert’s Paradox, and
showed that the early nineteenth-century work of Navier and Stokes, which generalized Newton’s law of friction, was indeed correct. The viscous theory of fluid motion advanced dramatically during the twentieth century with profound contributions from pioneers such as
Kolmogorov, Prandtl, Taylor and von Kármán.
During the second half of the twentieth century, a third branch of the unified theory of fluid
mechanics evolved, viz., Computational Fluid Dynamics, or CFD. Using digital computers,
we can now solve very complex fluid-flow problems. Early researchers such as Richardson,
Courant, Friedrichs, Lewy, Lax and von Neumann helped build the foundation of what has
become an important tool for both basic and applied research activities. The list of active
CFD researchers has expanded dramatically since the 1960’s, and many of the most significant
contributors in this rapidly developing field are currently expanding its horizons.
Figure 1.22: The personal computer (PC) of the early twenty-first century has computing
power more than triple that of a late 1980’s Cray Y-MP supercomputer. Capable of performing
in excess of nearly one billion floating-point operations per second for complex computer
programs, a PC with a 2.50-GHz Intel i5 four-core processor brings complex CFD capability
to the desktop of the engineer at very low cost. And, unlike the Cray Y-MP, it doesn’t require
liquid helium to cool any of its components!
38
CHAPTER 1. INTRODUCTION
Chapter Summary
Key topics discussed in this chapter...
• Dimensions and Units. Dimensions denote physical properties in a generic sense independent of units. Thus, we refer to dimensions of mass, length, temperature, time, etc.
By contrast, units are explicit scales for quantifying the magnitude of physical properties. There are two sets of units that find widespread use in fluid mechanics, namely,
Standard International (SI) units and the U. S. Customary System (USCS).
• Continuum Approximation. We treat a fluid as a continuum rather than a collection
of molecules. This approximation is valid as long as a fluid particle contains a large
number of molecules.
• Thermodynamic Properties. The science of fluid mechanics is a combination of
classical mechanics and thermodynamics. Key thermodynamic properties that appear in
fluid-mechanics applications are pressure, p, mass density, ρ, temperature, T , specific
internal energy, e, specific enthalpy, h = e + p/ρ, specific-heat coefficients, cv and cp ,
and specific-heat ratio, γ = cp /cv .
• Special Properties of Fluids. Other particularly interesting properties of fluids are
vapor pressure, pv , surface tension, σ, compressibility, τ , viscosity, μ, and kinematic
viscosity, ν.
• Fluid Types. A real fluid can support tangential stresses. An ideal, perfect or frictionless
fluid cannot support a tangential stress. Most of the material in the first 11 chapters
focuses on perfect fluids, while Chapters 12 through 15 cover real-fluid behavior.
Important equations introduced in this chapter...
• Perfect-Gas Law: Equations (1.25) and (1.27)
p = ρRT
or
p=
ρ RT
M
• Laplace’s Formula for Surface Tension: Equation (1.39)
σ
σ
+
= ∆p
R1 R2
• Capillary-Tube Rise Height: Equation (1.42)
∆h =
Pσ
cos φ
ρgA
• Shear Stress for a Newtonian Fluid: Equation (1.44)
τ =μ
du
dy
PROBLEMS
39
Problems
1.1 Using Tables 1.3 and 1.4, express 112 lbs, 55 mph and 309 ft in SI units.
1.2 Using Tables 1.3 and 1.4, express 218 Btu, 318 hp and 14.6 knots in SI units.
1.3 Using Tables 1.3 and 1.4, express 10 kg, 1050 Pa and 100 cm2 in USCS units.
1.4 Using Tables 1.3 and 1.4, express 74.6 kW, 126 cal and 28.4 L in USCS units.
1.5 Aviation pioneer Amelia Earhart weighed 116 lbs and stood 5 feet 8 inches tall. What was Earhart’s
mass in slugs? Determine her mass and height in kg and m, respectively.
Amelia Earhart
Howard Hughes
Babe Ruth
Roberto Clemente
Problems 1.5, 1.6, 1.7, 1.8
1.6 Aviation pioneer Howard Hughes weighed 150 lbs and stood 6 feet 4 inches tall. What was Hughes’
mass in slugs? Determine his mass and height in kg and m, respectively.
1.7 Baseball great Babe Ruth stood 1.88 m tall and his mass was 97.5 kg. What was his height and
weight in ft and lbs, respectively?
1.8 Baseball great Roberto Clemente stood 1.80 m tall and his mass was 79.4 kg. What was his height
and weight in ft and lbs, respectively?
1.9 In Old Russian units, a one-pound weight has a mass of 0.41 kg. What would the value of the
gravitational acceleration, gr , have to be for an object weighing 1 pound in Old Russian units to weigh
1 USCS pound with g = 32.174 ft/sec2 ?
1.10 An astronaut weighs 2.25 stone on the Moon. Noting that a stone is 14 pounds, what is the
astronaut’s weight, in pounds, on Earth where the gravitational acceleration is 6 times that of the Moon?
Problem 1.10
1.11 In research studies involving highly explosive substances, a research scientist has chosen to express
mass in kilograms (kg), time in milliseconds (msec) and force in kiloNewtons (kN). Beginning with the
definition of the Newton, 1 N = 1 kg·m/sec2 , infer the logical choice of units for length.
1.12 In research studies involving highly explosive substances, a 19th-century scientist chose to express
force in tons (ton), time in kiloseconds (ksec) and length in miles (mi). Beginning with the definition
of the pound, 1 lb = 1 slug·ft/sec2 , infer the logical choice of units for mass.
1.13 In order to study atomic particles, a scientist chooses to express mass in atomic mass units (amu,
1 kg = 6.023·1026 amu), force in picoNewtons (pN) and length in nanometers (nm). Beginning with the
definition of the Newton, 1 N = 1 kg·m/sec2 , infer the logical choice of units for time.
40
CHAPTER 1. INTRODUCTION
1.14 What force is required to impart an acceleration of 1 AU/yr2 to a 1 kiloslug mass? AU denotes
Astronomical Unit (93 · 106 miles) and yr denotes year (365 14 days).
1.15 The average distance traveled by a molecule between collisions is called the mean free path, mf p .
The continuum limit is valid if the characteristic√dimension of the flow is very large compared to mf p .
From the kinetic theory of gases, mf p = 1.5ν/ γRT , where ν is kinematic viscosity, γ is specific-heat
ratio, R is perfect-gas constant and T is absolute temperature. Compute the ratio of the diameter of
Imperial Standard 50 Gauge wire, d = 0.001 in, to the mean free path of carbon dioxide at 59o F. Does
flow of carbon dioxide at this temperature past this type of wire fall within the continuum limit?
1.16 The average distance traveled by a molecule between collisions is called the mean free path, mf p .
The continuum limit is valid if the characteristic√dimension of the flow is very large compared to mf p .
From the kinetic theory of gases, mf p = 1.5ν/ γRT , where ν is kinematic viscosity, γ is specific-heat
ratio, R is perfect-gas constant and T is absolute temperature. Compute the ratio of the diameter of
Imperial Standard 45 Gauge wire, d = 0.0071 cm, to the mean free path of oxygen at 15o C. Does flow
of oxygen at this temperature past this type of wire fall within the continuum limit?
1.17 At an altitude of 100 km, the average distance between air molecules is approximately 3 · 10−4 mm.
Compute the number of molecules contained in a spherical volume of diameter 1 μm.
1.18 Experience shows that a gas fails to behave as a continuum when it has fewer than 1012 molecules
per mm3 . If the temperature of air is 800o C, what is the lower bound on the pressure, in Pascals, for
the continuum limit to be valid? HINT: Write the perfect gas law as p = mnRT , where m is mass of
a molecule, n is number density, R is perfect-gas constant and T is absolute temperature.
1.19 How long would it take to count the molecules in 1 mm3 of air at STP with an extremely fast
electronic counter that can count at a rate of ten billion molecules per minute?
1.20 How long would it take to count the molecules in a cubic micron of water (1 micron is 10−4 cm)
with a very fast electronic counter that can count at a rate of three million molecules per second?
1.21 In a turbulent flow, the Kolmogorov length, η, is a measure of the smallest swirling motion, which
is commonly referred to as an eddy. For flow near the windshield of an automobile moving at 65 mph,
the Kolmogorov length is about 10−4 inch. Approximately how many air molecules are contained in a
spherical volume with diameter η?
1.22 When a typical amoeba swims in water, its shape can be approximated as a pair of spheres, each
with diameter d = 0.01 mm.
(a) How many water molecules are there in an amoeba-sized volume?
(b) If an amoeba considers a fluid particle to be a spherical volume with 10−3 d, how many molecules
are there in such a fluid particle?
Problem 1.22
1.23 If the pressure is 2200 psf and the temperature is 104o F, what is the ratio of the density of water
to the density of air, ρw /ρa ?
PROBLEMS
41
1.24 Given a temperature of T = 50o C and a pressure of p = 80 kPa, determine the density for air,
ammonia, helium, methane, nitrogen and oxygen.
1.25 Compute the weights, in Newtons, of 1 m3 of helium and of oxygen pressurized to 600 kPa and
held at a temperature of 80o C.
1.26 Compute the weights, in pounds, of 2 ft3 of methane and of hydrogen pressurized to 100 psi and
held at a temperature of −40o F.
1.27 Two pounds of methane are contained in a volume of 2 ft3 at a pressure of 6 atm. What is the
temperature?
1.28 One kilogram of nitrogen is contained in a volume of 200 liters at a temperature of −10o C. What
is the pressure?
1.29 What pressure (in atm) is required to compress air to a point where its density is 3% that of water
if the temperature is 40o C?
1.30 Find the effective molecular weight of air, where the units of molecular weight are slug/(slug-mole).
1.31 Using the value of R in Table A.1 for carbon dioxide, what is its molecular weight in kg/(kg-mole)?
Does this agree with what you learned in chemistry class?
1.32 Referring to the periodic tables in your chemistry book, and noting that the units of molecular
weight are kg/(kg-mole), determine the value of the perfect-gas constant, R, for methane, CH4 . How
does this compare with the value of R in Table A.1 for methane?
1.33 Assuming water vapor can be treated as a perfect gas, what is its gas constant, R, if, for a pressure
of 400 kPa and a temperature of 15o C, the density is 3 kg/m3 ?
1.34 Assuming water vapor can be treated as a perfect gas, what is its gas constant, R, if, for a pressure
of 8000 psf and a temperature of 80o F, the density is 5.35 · 10−3 slug/ft3 ?
1.35 Determine the specific weights of air and water, in N/m3 , at 20o C and 1 atm.
1.36 Determine the specific weights of air and water, in lb/ft3 , at 68o F and 1 atm.
1.37 Verify that, at 20o C and 1 atm, if the gas constant is R, then the specific gravity of a gas is
SG = Rair /R. Using this result, compute the specific gravity of CO2 , He, H2 , CH4 , N2 and O2 .
1.38 Compute the specific gravity of ethanol, seawater and mercury.
1.39 A dry fruit cake sparsely populated with cherries measures 3 in × 1.5 in × 6.8 in and weighs
16.9 oz. Determine the cake’s specific gravity relative to water at 68o F.
Problems 1.39, 1.40
1.40 A moist, cherry-laden fruit cake measures 7.6 cm × 4.3 cm × 15.2 cm and has mass 0.5 kg.
Determine the cake’s specific gravity relative to water at 20o C.
42
CHAPTER 1. INTRODUCTION
1.41 For a calorically-perfect gas, the specific-heat coefficients, cp and cv , and hence the specific-heat
ratio, γ = cp /cv , are all constant. For such a gas, derive expressions for cp and cv in terms of R and γ.
1.42 At high pressure and low temperature, the perfect-gas law is inaccurate. A better approximation is
provided by the van der Waal’s formula, which is
(p + aρ2 )(1 − bρ) = ρRT
where a = 27(RTc /8)2 /pc and b = (RTc /8)/pc . Determine the density, ρc , corresponding to pc and
Tc . HINT: Assume pc = N ρc RTc and solve for N , noting that the resulting cubic equation for N has
a triple root.
1.43 A closed tank contains liquid with open space above the liquid. All of the air has been removed.
(a) What will the pressure above the liquid be after several hours have passed if the liquid is gasoline
at a temperature of 16o C?
(b) If the liquid were water, at what temperature (to the nearest degree) would the same pressure be
realized?
1.44 A closed tank contains liquid with open space above the liquid. All of the air has been removed.
(a) What will the pressure above the liquid be after several hours have passed if the liquid is ethanol
at a temperature of 68o F?
(b) If the liquid were water, at what temperature (to the nearest degree) would the same pressure be
realized?
1.45 On a mountain-climbing expedition, climbers found that water boiled at 200o F. What was the
atmospheric pressure? Express your answer in atm.
1.46 On a mountain-climbing expedition, climbers found that water boiled at 94o C at their first camp
site. At the top of the mountain, they noted that water boiled at 83o C. How much lower was the
atmospheric pressure at the top relative to its value at the first camp site? Express your answer in atm.
1.47 A researcher has designed a test rig to demonstrate the effect of temperature on cavitation. In the
test section of the test rig, the pressure varies as
p = pa 1 −
7
8
s
c
2
where p is test-section pressure, pa is atmospheric pressure, s is distance from the beginning of the test
section and c is test-section length. The fluid used in the experiment is water. If cavitation is observed
at s/c = 0.6, determine the temperature.
1.48 The pressure near a 90o elbow of a piping system has been found by experiment to be well
approximated by
p = pa − λpa x/R
where pa is atmospheric pressure, x is distance from a point just upstream of the elbow, R is the radius
of curvature and λ is a dimensionless constant. The fluid is water.
(a) If the temperature is 20o C and cavitation occurs when x = R, what is the value of λ?
(b) Using the results of Part (a) for λ, at what value of x/R will cavitation occur if the temperature
increases to 60o C?
1.49 The pressure, p, near a pipe inlet can be estimated from a curve fit to measurements, which is
p = pa e−s/so
where pa is atmospheric pressure, s is arclength and so = 2 m. The fluid is water at 70o C.
(a) At what value of s will cavitation occur?
(b) To postpone the onset of cavitation to s = 5 m, what must the temperature be?
PROBLEMS
43
1.50 What pressure change, ∆p, starting from atmospheric pressure, is required to cause a 1% change
in density for helium, mercury and water? Assume the process is isothermal, and express your answers
in MPa (megaPascals).
1.51 At a pressure of 500 psi, a sample of water occupies a volume of 1 ft3 . When the pressure
increases to 4000 psi, the volume decreases by 1%. Estimate the value of the compressibility, τ , under
these conditions, assuming the compressibility of water remains nearly constant.
1.52 What pressure change, ∆p, starting from atmospheric pressure, is required to cause a 5% change
in density for air, water and glycerin? Assume T = 60o F, and express your answers in atm.
1.53 Assuming temperature is 15.6o C, compute the fractional change in density, ∆ρ/ρ, for air and water
caused by a change in pressure, ∆p = 0.35 atm. The initial pressure is p = 1 atm.
1.54 When the pressure increases from 14.7 psi to 19 psi, what is the fractional change in density, ∆ρ/ρ,
for air? Assume constant temperature.
1.55 At very high pressures, the compressibility of water is nearly constant. If the pressure on 1 liter of
water increases from 60 atm to 600 atm, what is the change in volume?
1.56 An empirical formula relating pressure and density for water with temperature held constant is
p/pa ≈ 3041(ρ/ρa )n − 3040
where pa and ρa are conditions at the surface and n is a dimensionless constant.
(a) According to this formula, what is the compressibility, τ , of water as a function of p, pa and n?
(b) Appealing to Equation (1.36), determine the numerical value of the constant n.
1.57 An empirical formula relating pressure and density for seawater with temperature held constant is
p/pa ≈ (α + 1)(ρ/ρa )7 − α
where pa and ρa are conditions at the surface and the dimensionless constant α is approximately 3000.
The pressure at the deepest part of the Pacific Ocean is about 1100 atm. What is the density of seawater
at this depth in kg/m3 ? What is the percentage difference from the value at the surface?
1.58 If a gas is compressed through a frictionless process with no heat exchange, the pressure is related
to density by the isentropic relation, p = Aργ , where A is a constant and γ is the specific-heat ratio.
(a) Compute the compressibility, τs , for such a process. Express your answer in terms of γ and p.
(b) Compare your result with the isothermal τ for a gas when p = 1 atm.
1.59 A spherical soap bubble has an inside radius, R = 8 mm, film thickness, t
R, and surface
tension σ. Noting that a bubble has two liquid-air interfaces, find the difference between internal and
ambient pressure, ∆p, as a function of σ and R. Assuming σ = 0.073 N/m, what is ∆p?
1.60 A child is blowing bubbles using an elliptical ring. At a point during formation of a bubble, the
principal radii of curvature for the soap film are R1 = 54 R and R2 = λR, where λ is a constant. If the
pressure difference that the film can support, ∆p, is identical to that for a sphere of radius R, what is
the value of λ?
1.61 Air surrounds a water droplet of diameter D = 0.012 in. The pressure within the droplet is
∆p = 19 psf above ambient. Determine the surface tension and the temperature.
1.62 Air surrounds a water droplet of diameter D = 20 mm. The pressure within the droplet is
∆p = 15 Pa above ambient. Determine the surface tension and the temperature.
44
CHAPTER 1. INTRODUCTION
1.63 A small droplet of mercury is immersed in air. Compute the ratio of its diameter to the diameter it
would have when immersed in water. Assume the pressure within the droplet exceeds the ambient value
by the same amount in both cases.
1.64 A small droplet of ethanol is immersed in air. Compute the ratio of its diameter to the diameter
it would have when immersed in mercury. Assume the pressure within the droplet exceeds the ambient
value by the same amount in both cases.
1.65 Compare the pressure difference between the interior and exterior of a water drop, ∆pH2 O , to that
of a drop of mercury from a broken thermometer, ∆pHg . The water drop is 1/3 cm in diameter and the
drop of mercury is 1/2 cm in diameter. The drops are spherical and the air temperature is 20o C.
1.66 To what height, ∆h, will mercury rise in a cylindrical capillary tube of diameter, d = 1/3 in?
Assume the wetting angle is φ = 129o .
.
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−
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σ
ª
∆h
..
d
..
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.
g
...
............. ....
..
... ........
..........................................................................................................
ρ
Problems 1.66, 1.67, 1.68, 1.69, 1.70
1.67 To what height, ∆h, will glycerin rise in a cylindrical capillary tube of diameter, d = 0.01 in?
Assume the wetting angle is φ = 18o .
1.68 To what height, ∆h, will ethanol rise in a cylindrical capillary tube of diameter, d = 4 mm?
Assume the wetting angle is φ = 0o .
1.69 Suppose a square tube of width d is immersed in a large container. If the fluid density is ρ, the
surface tension is σ and the wetting angle is φ, to what height, ∆h, will the fluid rise?
1.70 Suppose a rectangular tube d wide by 3d thick out of the page is immersed in a large container. If
the fluid density is ρ, the surface tension is σ and the wetting angle is φ, to what height, ∆h, will the
fluid rise?
1.71 Two tubes are used in a laboratory experiment to demonstrate capillary action. One tube has square
cross section with side s. The other tube is cylindrical with radius r. Assume the wetting angle is the
same for both tubes.
(a) Determine the value of s as a function of r required for the fluid to rise to the same level, h, in
both tubes.
(b) Find the ratio of the volume of fluid in the square tube, Vsq , to the volume of fluid in the
cylindrical tube, Vcyl . Explain why Vsq > Vcyl .
1.72 An open tube of diameter d is inserted into a pan of carbon tetrachloride. A second tube of diameter
2d is inserted into a pan of water. Due to capillary action, the fluids rise to heights of ∆h1 and ∆h2
for the carbon tetrachloride and water, respectively. Compute the ratio ∆h2 /∆h1 . Assume the wetting
angle is the same in both tubes and T = 20o C.
1.73 A round capillary tube of diameter d is inserted into a pan of water. A second tube, also of diameter
d, is inserted into a pan of unknown liquid whose density is 32 ρw , where ρw is the density of water. If
4
the liquid rises to a height ∆h in water and 15
∆h in the unknown liquid, what is the surface tension
of the unknown liquid? Assume the wetting angle is the same in both tubes and that T = 68o F.
PROBLEMS
45
1.74 A small U-shaped capillary tube with square cross section of side s is inserted in adjacent pans of
liquid as shown in the figure. The fluid rises to its equilibrium heights of ∆h1 and ∆h2 in the left and
right segments, respectively. Fluid density and surface tension in the left pan are ρ and σ, respectively.
Fluid density and surface tension in the right pan are 53 ρ and 13 σ, respectively. The wetting angle for
the fluid on the right side of the tube is 0o . The air in the tube remains at atmospheric pressure. If
∆h1 = ∆h2 , what is the wetting angle for the fluid in the left side of the tube, φ?
s
..................................................................................................................................
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∆h1
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...................................................... ...........................................................
s
...
...
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..
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.....
.
...
.
.....................................................................
g
5
ρ, 13 σ
3
ρ, σ
Problem 1.74
1.75 A small cylindrical capillary tube inserted in a pan of water has radius r. After the fluid rises in
the tube to its equilibrium height, it is closed at the top and the pressure of the air trapped in the tube
is increased to pb . Assume that the wetting angle, φ, is 0o , surface tension is σ and water density is ρ.
Also, g is gravitational acceleration and pa is atmospheric pressure.
(a) Determine the value of ∆h as a function of σ, ρ, g, r, pa and pb .
(b) Now, assuming σ = 2rpa , what is the ratio of pb to pa required to make ∆h half of what it
would be if the capillary tube were open at the top?
...
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...
b
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p
−
ª
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∆h
..
g
...
...
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a
... ........
.
.........................................................................................................
............
a
.
...
........................................................................................................
2r............
p
σ
p
ρ
Problem 1.75
1.76 A small cube of side s is floating in a liquid of density ρ. One fifth of the cube is submerged.
The density of the cube is 35 ρ and the surface tension between the cube and the liquid is σ. Also, the
wetting angle is φ = 0o .
(a) What is the pressure on the bottom face of the cube? Express your answer in terms of ρ, s,
atmospheric pressure, pa , and gravitational acceleration, g.
(b) There are four vertical forces acting on the cube. Name them and compute the value of each in
terms of the given information.
(c) Balancing forces, determine the cube’s side, s, as a function of σ, ρ and g.
.
......
..
..
s.
...............................................................................
........ ..
.
................
ρ
...
...
.
...........
.....
.
3
ρ
5
s
g
..
.......
..........................................................
..................1
..
.
...... 5
................ ....
Problem 1.76
s
46
CHAPTER 1. INTRODUCTION
1.77 A piece of wire of radius r and length = 16r is floating on the surface of a liquid of density ρ.
The wire’s density is ρw = 32 ρ and it is exactly half submerged. The wetting angle between the wire
and the liquid is φ = 0o . There are three forces acting on the wire, viz., its weight, surface tension and
buoyancy. The buoyancy force is Fb = 8πρgr3 upward. Compute r as a function of ρ, gravitational
acceleration, g, and surface-tension coefficient, σ.
...
...
g = g k.........
.......
.....
¢
¢
¢
¢
¢ ......... .......
ρ
.....
............
.... ...
w
. .............................................................................................
........ .......................................... .........................................
.
..
.
ρ
2r
¢
¢
¢ ¢¢
¢¢
¢¢
¢
x
.
..........
...
................................
..
..........
..
y
z
Problem 1.77
1.78 A small spherical ball of diameter d is floating on the surface of a liquid. The ball is half submerged
and its weight, Fw , is supported by surface tension, Fσ , and the buoyancy force, Fbuoy . The buoyancy
force acts upward with magnitude 12 ρgV , where V = 16 πd3 is the volume of the sphere, ρ is the density
of water and g is the acceleration of gravity. The density of the sphere is ρs and surface tension is σ.
(a) Determine the ratio of the sphere’s density to that of water, ρs /ρ, as a function of σ, ρ, g, d and
wetting angle, φ.
(b) At 68o F, the surface tension and density of water are σ = 0.005 lb/ft and ρ = 1.94 slug/ft3 .
Assuming the wetting angle is φ = 0o relative to the vertical direction, what is the density ratio,
ρs /ρ, for a sphere of diameter d = do = 19 in?
√
(c) From your solution of Part (a), if ρs , σ and ρ are unchanged, verify that d = do cos φ, where
do is the diameter corresponding to zero wetting angle.
(d) If the wetting angle is 20o , use your results of Parts (b) and (c) to determine the sphere’s diameter.
.
w .....
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F
φ
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Fσ
·
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g
d
ρ
¹ ........ ¸
....
....
Fbuoy
Problem 1.78
1.79 Consider a liquid-air interface next to a plane wall. As illustrated in the figure, surface tension
causes the liquid to assume the shape z = η(x). The surface tension is σ, the fluid density is ρ and
the wetting angle is φ. You can assume the pressure difference across the interface and the radius of
curvature of the interface are
1
d2 η
∆p ≈ ρgη
and
≈
R
dx2
where g is gravitational acceleration. Determine the shape of the interface and the maximum height, h.
z ...........
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z=h
φ
g = −g k
z = η(x)
x
Problem 1.79
PROBLEMS
47
1.80 Plane Poiseuille flow corresponds to flow between two infinite parallel plates. The velocity distribution, u(y), is given by
y
y
1−
u(y) = um
h
h
where um is a constant velocity scale, y is measured from the lower wall and the space between the
walls is 2h. Plot the velocity profile as u/um versus y/h, and determine the shear stress at both walls
as a function of μ, um and h.
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y ........
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.......
....
..
.
u(y)
x
2h
..
...
...
......
.
Problem 1.80
1.81 Viscous flow close to a solid surface develops as shown in the figure. The region 0 ≤ y ≤ δ is
called the boundary layer, δ is the boundary-layer thickness and Ue is the boundary-layer edge velocity.
Determine the surface shear stress (the value at y = 0) for the following velocity profiles.
y
(a) u(y) = Ue
δ
πy
(b) u(y) = Ue sin
2δ
(c) u(y) = Ue
y
δ
3
1
−
2
2
y
δ
y
2
Ue
..
......
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. .. .. .. .. ..
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...................................... ....
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....
δ
U
......
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x
Problem 1.81
1.82 We want to approximate the viscosity of hydrogen with a power law, i.e., μ = μr T ω . Using
Sutherland’s law (Table A.8) to determine μ when T = 200 K and 1000 K, compute μr and ω.
1.83 In many mid-20th century research studies, the viscosity of air was approximated by a power law
according to μ = AT 0.7 , where T is absolute temperature and A is a constant.
(a) Determine A by making the value of μ at T = 68o F equal the value given by Sutherland’s law.
(b) Make a graph of μ according to the power law developed in Part (a) and Sutherland’s law for
−250o F ≤ T ≤ 2000o F.
1.84 The viscosity of water as a function of temperature can be represented by Andrade’s equation,
viz., μ = μr eTr /T . The empirical Sutherland equation, Equation (1.47), is often used to compute
the viscosity of air. In both cases, the quantities A, S, Tr and μr are empirical coefficients that can
be determined from measurements—all four are positive numbers. Using these formulas and assuming
constant pressure, determine the sign of dν/dT for water and for air, where ν is kinematic viscosity.
1.85 The viscosity of water can be approximated by Andrade’s equation, viz., μ = μr eTr /T , where μr
and Tr are empirical viscosity and temperature scales, respectively, and T is absolute temperature.
(a) Using values from Tables A.3 and A.7 for T = 10o C and T = 90o C, determine μr and Tr .
(b) Using the values of μr and Tr determined in Part (a), make a graph comparing the tabulated
values of μ with Andrade’s equation for 0o C ≤ T ≤ 100o C.
48
CHAPTER 1. INTRODUCTION
1.86 Human blood is a non-Newtonian fluid whose shear stress can be approximated by τ ≈ K(du/dy)n
where K and n are given in Table A.10.
(a) Determine the effective viscosity, μe , such that τ = μe du/dy.
(b) Compute the ratio of μe to the viscosity of water, μw , at 20o C.
(c) Make a graph of μe /μw for values of du/dy between 0.01 sec−1 and 10 sec−1 .
1.87 Heavy cream is a non-Newtonian fluid whose shear stress can be approximated by τ ≈ K(du/dy)n
where K and n are given in Table A.10.
(a) Determine the effective viscosity, μe , such that τ = μe du/dy.
(b) Compute the ratio of μe to the viscosity of water, μw , at 20o C.
(c) Make a graph of μe /μw for values of du/dy between 0.01 sec−1 and 10 sec−1 .
1.88 In an experimental study of non-Newtonian fluids, a neutrally buoyant circular cylinder of radius
R and length slides inside a vertical smooth pipe with inside radius R + h, where h
R. The
pipe is filled with a fluid whose shear stress, τ , varies with velocity gradient, du/dy, according to
τ = K (du/dy)n , where y is distance measured from the pipe surface inward and the quantities K and
n are constants. A cable pulls the cylinder slowly upward at constant speed, U . The cable tension,
T (U ), is measured for three different fluids, and we wish to determine the percentage change in cable
tension when the speed is doubled, i.e., η = 100[T (2U ) − T (U )]/T (U ). Because the cylinder is
neutrally buoyant, the cable-tension force equals the viscous force on the cylinder.
(a) Couette flow exists between the cylinder and pipe surfaces, and the exact viscous-flow equations
show that τ is constant throughout the gap. Determine du/dy as a function of τ , K and n, solve
for u(y), and develop an equation for τ in terms of K, n, h and U .
(b) Solve for η in terms of K, n, R, h and U .
(c) Compute η for ketchup, water and quicksand.
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•
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....
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.... .
................. . .....................
...................................
...............
T
R+h
R
U
Problem 1.88
1.89 A plate moving with velocity U in its own plane lies between two stationary parallel plates as
shown with m = 2. The viscosity of the fluid below the moving plate is μ, while the fluid above has
viscosity μu = 4μ. Make a graph of the velocity, u/U , versus y/h for 0 ≤ y/h ≤ 3 and compute the
shear stress on the upper and lower walls.
y .........
....
....
...
..
..................................
mh
x
h
.
.......
..
..
.......
.......
.......
.
μu
μ
U
.......................................................
Problems 1.89, 1.90
1.90 A plate moving with velocity U in its own plane lies between two stationary parallel plates as
shown. What must the viscosity of the fluid above the moving plate be, as a function of μ and m, in
order for the magnitude of the shear stress above the plate to be one fourth the value below?
PROBLEMS
49
1.91 A rotating-cylinder viscometer can be used to measure the viscosity of a liquid. This device consists
of a cylinder of radius Ri and length L rotating at angular velocity Ω inside a concentric cylinder of
radius Ro . The device is designed to have Ro − Ri
Ri and Ro − Ri
L. Determine the torque,
T , on the rotating cylinder as a function of Ri , Ro , L, Ω and μ. Neglect any torque on the bottom of
the inner cylinder.
........
.. .
........... .... .... ........................
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.........
....................................................
.... ...
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... ... ...
... .... ...
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i
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... ... ...
... .. ..
...................................................................... .... .... ...
... ... ............................
......... .. ..
... ............................................................................................................. ....
........................................................................................................................
Ω
.
.......
...
..
2R
............................................
2Ro
L
...
...
.......
.
............................................
Problem 1.91
1.92 Compute the average velocity, u, and the average specific kinetic energy,
defined by
h
h
1
1
1 2
1 2
u≡
u(y)dy
and
u ≡
u (y)dy
2
h 0
h 0 2
1 2
u ,
2
for Couette flow
U
¡
..........................................................
y
...
.......
....
...
..
...
..
...........................................
........................................
¡
¡
¡
¡
h.
...
...
...
...
..
.......
.
........................
¡
x
....................................................
.
.......
..
...
...
...
..
Problems 1.92, 1.93, 1.94
1.93 Measurements show that the shear stress for Couette flow is τ = 40 Pa. The plate-separation
distance is h = 1 mm and the velocity is U = 10 m/sec.
(a) What is the viscosity of the fluid between the plates?
(b) What is the maximum density for which the answer of Part (a) is valid?
1.94 Measurements show that the shear stress for Couette flow is τ = 0.18 lb/ft2 . The plate-separation
distance is h = 2 in and the velocity is U = 12 ft/sec.
(a) What is the viscosity of the fluid between the plates?
(b) What is the maximum density for which the answer of Part (a) is valid?
1.95 Insulating paint is applied to a wire by pulling it through a cylindrical orifice of radius R. The
wire, of radius r, is centered in the orifice whose length is . The viscosity of the paint is μ. What
force, F , is required to pull the wire at a constant velocity, U ? Assume (R − r)
R and (R − r)
.
lubricant, μ ............................
........
... ...
..... ...
... ...
... ...
.....
..........
... ...........
......
.. ..
..
......
....
.
...............
...
..
........
.
2R
.......
..
............................................................
......
...
...
..
...
..
.
.
.
.
.
...
..
.......
.
.
........
..
2r
.....
.. ...
.. ..
. .....................................
. .. .
.. ..
... ...
..
U
............................................................
Problem 1.95
1.96 The platter in a standard 3.5-inch hard-disk drive rotates at Ω = 7200 rpm. The platter diameter, D,
is 3.74 in and the read-write head flies h = 50 nanoinches from the platter. Assuming the pressure in
the drive is atmospheric and the temperature is 100o F, can flow between the head and the platter be
approximated as Couette flow?
50
CHAPTER 1. INTRODUCTION
1.97 A circular cylinder of radius r, length and weight W slides inside a vertical pipe having an inside
radius R. A frictionless guide wire maintains the cylinder’s orientation. The small space between the
cylinder and pipe is lubricated with a thin film that has viscosity μ. Derive a formula for the acceleration,
dU/dt, of the cylinder in the vertical pipe in terms of r, R, , μ and W . Use the formula to find the
terminal velocity of the cylinder (the velocity when dU/dt → 0) for r = 5.97 in, R = 6 in, = 12 in
and W = 3 lb. Determine the terminal velocity for water at 50o F, SAE 10W Oil at 100o F and glycerin
at 68o F. For which of these three cases is the Couette-flow solution valid?
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Problem 1.97
1.98 A cylinder of mass m slides horizontally in a lubricated, but otherwise empty, pipe. A frictionless
guide wire (not shown) maintains the cylinder’s orientation. The radial clearance between cylinder and
pipe is 12 d. The diameter and length of the cylinder are D and L, respectively. Assuming the cylinder’s
deceleration is −a when the speed is U , derive a formula for the viscosity of the fluid in terms of m,
a, d, D, L and U . If the weight of the cylinder is 100 N, d = 0.05 mm, a = 1 m/sec2 , U = 7 m/sec,
D = 15 cm and L = 8 cm, determine the value of μ in kg/(m·sec).
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Problem 1.98
1.99 A cube of density ρ and side s slides down an inclined plane coated with a thin lubricating film of
thickness h and viscosity μ. Assuming the Couette-flow solution holds in the lubricating film, derive a
differential equation for the motion of the cube. Solve the differential equation, assuming the cube starts
from rest at time t = 0. What is the terminal velocity for ρ = 19300 kg/m3 , s = 0.04 m, h = 0.001 m,
μ = 1.60 kg/(m·sec) and α = 30o ?
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Problems 1.99, 1.100, 1.101
1.100 A cube of density ρ and side s slides down an inclined plane coated with a thin lubricating film
of thickness h and viscosity μ. Its terminal velocity is U . Assume the Couette-flow solution holds in
the lubricating film.
(a) Develop a formula for the thickness of the lubricating film.
(b) Compute h for ρ = 7800 kg/m3 , s = 0.05 m, μ = 0.068 kg/(m·sec), U = 20 m/sec and α = 45o .
(c) What is the maximum density of the fluid in the lubricating film for which the solution is valid?
PROBLEMS
51
1.101 A cylinder of density ρ, diameter s and height s slides down an inclined plane coated with a
thin lubricating film of thickness h and viscosity μ. Assuming the Couette-flow solution holds in the
lubricating film, derive a differential equation for the motion of the cylinder. Solve the differential
equation, assuming the cylinder starts from rest at time t = 0. What is the terminal velocity for
ρ = 37.4 slug/ft3 , s = 2 in, h = 0.02 in, μ = 0.0023 slug/(ft·sec) and α = 15o ?
1.102 A block of mass M slides on a thin film of oil whose viscosity is μ. The film thickness is h
and the area of the block exposed to the oil is A. Another block of mass m is attached with a cord as
shown. When released, the mass m exerts tension on the cord, causing the blocks to accelerate. You
may neglect friction in the pulley and air resistance. Assuming the Couette-flow solution holds in the
lubricating film, and that the instantaneous speed of the block is U (t), derive a differential equation for
U as a function of time. Obtain an algebraic expression for the maximum speed of the block. HINT:
Draw free-body diagrams for each mass with the tension force acting on each.
U
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Problem 1.102
1.103 Two cubes are arranged on an incline with a pulley as indicated. The objects are free to slide
on thin oil films of different thicknesses with the same viscosity, μ. Using the notation indicated in the
figure, set up the differential equation of motion for the system. Assume the Couette-flow solution holds
in the lubricating film, and that h1
s1 and h2
s2 .
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Problem 1.103
1.104 You were bored waiting for commercials to scan by while watching your favorite documentary
on fluid mechanics. You decided to fix your VCR motor, increasing the scanning speed for a 6-hour
tape. In Super Long Play (SLP) mode, 3 minutes of commercials now scan in 12.5 seconds instead of
50 seconds. Compute the tape speed, U , before and after the alteration. By what factor did you increase
the tension in the tape as it passes by the head? Assume the Couette-flow solution holds in the small
gap between the tape and the head. A standard 6-hour tape is 807 feet long.
.
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Problem 1.104
1.105 Compute the average specific kinetic energy for pipe flow, 12 u2 , defined by
1 2
u
2
≡
1
πR2
2π
R
1 2
u (r)rdrdθ
2
0
0
52
CHAPTER 1. INTRODUCTION
1.106 We wish to analyze flow in a pipe of radius R.
(a) Compute the average velocity for pipe flow, u, defined by
u≡
1
πR2
2π
R
u(r)rdrdθ
0
0
(b) Derive an equation for the friction factor, f , as a function of the Reynolds number based on pipe
diameter, ReD , according to the following definitions.
f≡
−τ (R)
1
2
8 ρu
and
ReD ≡
ρuD
μ
1.107 Derive an equation for the skin friction, cf , as a function of Reynolds number based on pipe
radius, ReR , for pipe flow, where R denotes pipe radius. By definition,
cf ≡
−τ (R)
1
2
2 ρum
and
ReR ≡
ρum R
μ
1.108 A defective pump moves water at 50o F through a small tube of diameter D = 0.24 in and length
L = 10 in. It creates a suction-pressure difference of p1 − p2 = 0.02 psi.
(a) What is the flow speed, um , at the center of the tube’s cross section?
(b) What is the Reynolds number, ReR ≡ um R/ν, where R is radius and ν is kinematic viscosity?
(c) What is the total viscous force acting on the tube in pounds?
1.109 A defective pump moves water at 20o C through a small tube of diameter D = 8 mm and length
L = 1 m. It creates a suction-pressure difference of p1 − p2 = 100 Pa.
(a) What is the flow speed, um , at the center of the tube’s cross section?
(b) What is the Reynolds number, ReR ≡ um R/ν, where R is radius and ν is kinematic viscosity?
(c) What is the total viscous force acting on the tube in Newtons?
1.110 Using the pipe-flow solution, we can analyze laminar flow of blood in the aorta.
(a) Show that, for a given pressure difference, the volume-flow rate, Q, increases as R4 , where
R
Q ≡ 2π 0 u(r)rdr.
(b) Now, write your answer in terms of the maximum velocity, um , and the radius, R. The maximum
velocity in the aorta is um ≈ 30 cm/sec and the radius is R ≈ 1.5 cm. How long does it take
for 200 cm3 of blood to flow through the aorta?
1.111 Using the pipe-flow solution, we can analyze laminar flow of blood in the aorta.
(a) Show that, for a given pressure difference, the volume-flow rate, Q, increases as R4 , where
R
Q ≡ 2π 0 u(r)rdr.
(b) Now, write your answer in terms of the maximum velocity, um , and the radius, R. The maximum
velocity in the aorta is um ≈ 1 ft/sec and the radius is R ≈ 0.6 in. How long does it take for
10 in3 of blood to flow through the aorta?
1.112 Water is flowing very slowly through a drain pipe of diameter D = 5 cm and length L = 120 m.
The pressure difference between inlet and outlet is (p1 − p2 ) = 0.04 kPa, and the temperature is 50o C.
Compute the maximum velocity and the Reynolds number of the water flowing in the pipe. Do you
expect the flow to be laminar or turbulent?
1.113 Water is flowing very slowly through a drain pipe of diameter D = 5 cm and length L = 120 m.
The pressure difference between inlet and outlet is (p1 − p2 ) = 0.04 kPa, and the temperature is 20o C.
Compute the maximum velocity and the Reynolds number of the water flowing in the pipe. Do you
expect the flow to be laminar or turbulent?